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What do creating a videogame, piloting a spaceship, and building a skyscraper all have in common? They all utilize geometry. Geometry is Professor Mary Pyo's favorite subject and she brings her excitement when she focuses on student problem areas such as proofs. Mary begins each lesson with fundamental concepts and reinforces them with many examples. Additional topics include Inductive/Deductive Reasoning, Congruency, Proportions, and Transformational Geometry. Professor Pyo received her Master’s in Educational Curriculum and Instruction, her specialized credentials in Foundational mathematics, and has been teaching for over 10 years.

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I. Tools of Geometry
  Coordinate Plane 16:41
   Intro 0:00 
   The Coordinate System 0:12 
    Coordinate Plane: X-axis and Y-axis 0:15 
    Quadrants 1:02 
    Origin 2:00 
    Ordered Pair 2:17 
   Coordinate Plane 2:59 
    Example: Writing Coordinates 3:01 
   Coordinate Plane, cont. 4:15 
    Example: Graphing & Coordinate Plane 4:17 
    Collinear 5:58 
   Extra Example 1: Writing Coordinates & Quadrants 7:34 
   Extra Example 2: Quadrants 8:52 
   Extra Example 3: Graphing & Coordinate Plane 10:58 
   Extra Example 4: Collinear 12:50 
  Points, Lines and Planes 17:17
   Intro 0:00 
   Points 0:07 
    Definition and Example of Points 0:09 
   Lines 0:50 
    Definition and Example of Lines 0:51 
   Planes 2:59 
    Definition and Example of Planes 3:00 
   Drawing and Labeling 4:40 
    Example 1: Drawing and Labeling 4:41 
    Example 2: Drawing and Labeling 5:54 
    Example 3: Drawing and Labeling 6:41 
    Example 4: Drawing and Labeling 8:23 
   Extra Example 1: Points, Lines and Planes 10:19 
   Extra Example 2: Naming Figures 11:16 
   Extra Example 3: Points, Lines and Planes 12:35 
   Extra Example 4: Draw and Label 14:44 
  Measuring Segments 31:31
   Intro 0:00 
   Segments 0:06 
    Examples of Segments 0:08 
   Ruler Postulate 1:30 
    Ruler Postulate 1:31 
   Segment Addition Postulate 5:02 
    Example and Definition of Segment Addition Postulate 5:03 
   Segment Addition Postulate 8:01 
    Example 1: Segment Addition Postulate 8:04 
    Example 2: Segment Addition Postulate 11:15 
   Pythagorean Theorem 12:36 
    Definition of Pythagorean Theorem 12:37 
   Pythagorean Theorem, cont. 15:49 
    Example: Pythagorean Theorem 15:50 
   Distance Formula 16:48 
    Example and Definition of Distance Formula 16:49 
   Extra Example 1: Find Each Measure 20:32 
   Extra Example 2: Find the Missing Measure 22:11 
   Extra Example 3: Find the Distance Between the Two Points 25:36 
   Extra Example 4: Pythagorean Theorem 29:33 
  Midpoints and Segment Congruence 42:26
   Intro 0:00 
   Definition of Midpoint 0:07 
    Midpoint 0:10 
   Midpoint Formulas 1:30 
    Midpoint Formula: On a Number Line 1:45 
    Midpoint Formula: In a Coordinate Plane 2:50 
   Midpoint 4:40 
    Example: Midpoint on a Number Line 4:43 
   Midpoint 6:05 
    Example: Midpoint in a Coordinate Plane 6:06 
   Midpoint 8:28 
    Example 1 8:30 
    Example 2 13:01 
   Segment Bisector 15:14 
    Definition and Example of Segment Bisector 15:15 
   Proofs 17:27 
    Theorem 17:53 
    Proof 18:21 
   Midpoint Theorem 19:37 
    Example: Proof & Midpoint Theorem 19:38 
   Extra Example 1: Midpoint on a Number Line 23:44 
   Extra Example 2: Drawing Diagrams 26:25 
   Extra Example 3: Midpoint 29:14 
   Extra Example 4: Segment Bisector 33:21 
  Angles 42:34
   Intro 0:00 
   Angles 0:05 
    Angle 0:07 
    Ray 0:23 
    Opposite Rays 2:09 
   Angles 3:22 
    Example: Naming Angle 3:23 
   Angles 6:39 
    Interior, Exterior, Angle 6:40 
    Measure and Degrees 7:38 
   Protractor Postulate 8:37 
    Example: Protractor Postulate 8:38 
   Angle Addition Postulate 11:41 
    Example: Angle addition Postulate 11:42 
   Classifying Angles 14:10 
    Acute Angle 14:16 
    Right Angles 14:30 
    Obtuse Angle 14:41 
   Angle Bisector 15:02 
    Example: Angle Bisector 15:04 
   Angle Relationships 16:43 
    Adjacent Angles 16:47 
    Vertical Angles 17:49 
    Linear Pair 19:40 
   Angle Relationships 20:31 
    Right Angles 20:32 
    Supplementary Angles 21:15 
    Complementary Angles 21:33 
   Extra Example 1: Angles 24:08 
   Extra Example 2: Angles 29:06 
   Extra Example 3: Angles 32:05 
   Extra Example 4 Angles 35:44 
II. Reasoning & Proof
  Inductive Reasoning 19:00
   Intro 0:00 
   Inductive Reasoning 0:05 
    Conjecture 0:06 
    Inductive Reasoning 0:15 
   Examples 0:55 
    Example: Sequence 0:56 
    More Example: Sequence 2:00 
   Using Inductive Reasoning 2:50 
    Example: Conjecture 2:51 
    More Example: Conjecture 3:48 
   Counterexamples 4:56 
    Counterexample 4:58 
   Extra Example 1: Conjecture 6:59 
   Extra Example 2: Sequence and Pattern 10:20 
   Extra Example 3: Inductive Reasoning 12:46 
   Extra Example 4: Conjecture and Counterexample 15:17 
  Conditional Statements 42:47
   Intro 0:00 
   If Then Statements 0:05 
    If Then Statements 0:06 
   Other Forms 2:29 
    Example: Without Then 2:40 
    Example: Using When 3:03 
    Example: Hypothesis 3:24 
   Identify the Hypothesis and Conclusion 3:52 
    Example 1: Hypothesis and Conclusion 3:58 
    Example 2: Hypothesis and Conclusion 4:31 
    Example 3: Hypothesis and Conclusion 5:38 
   Write in If Then Form 6:16 
    Example 1: Write in If Then Form 6:23 
    Example 2: Write in If Then Form 6:57 
    Example 3: Write in If Then Form 7:39 
   Other Statements 8:40 
    Other Statements 8:41 
   Converse Statements 9:18 
    Converse Statements 9:20 
   Converses and Counterexamples 11:04 
    Converses and Counterexamples 11:05 
    Example 1: Converses and Counterexamples 12:02 
    Example 2: Converses and Counterexamples 15:10 
    Example 3: Converses and Counterexamples 17:08 
   Inverse Statement 19:58 
    Definition and Example 19:59 
   Inverse Statement 21:46 
    Example 1: Inverse and Counterexample 21:47 
    Example 2: Inverse and Counterexample 23:34 
   Contrapositive Statement 25:20 
    Definition and Example 25:21 
   Contrapositive Statement 26:58 
    Example: Contrapositive Statement 27:00 
   Summary 29:03 
    Summary of Lesson 29:04 
   Extra Example 1: Hypothesis and Conclusion 32:20 
   Extra Example 2: If-Then Form 33:23 
   Extra Example 3: Converse, Inverse, and Contrapositive 34:54 
   Extra Example 4: Converse, Inverse, and Contrapositive 37:56 
  Point, Line, and Plane Postulates 17:24
   Intro 0:00 
   What are Postulates? 0:09 
    Definition of Postulates 0:10 
   Postulates 1:22 
    Postulate 1: Two Points 1:23 
    Postulate 2: Three Points 2:02 
    Postulate 3: Line 2:45 
   Postulates, cont.. 3:08 
    Postulate 4: Plane 3:09 
    Postulate 5: Two Points in a Plane 3:53 
   Postulates, cont.. 4:46 
    Postulate 6: Two Lines Intersect 4:47 
    Postulate 7: Two Plane Intersect 5:28 
   Using the Postulates 6:34 
    Examples: True or False 6:35 
   Using the Postulates 10:18 
    Examples: True or False 10:19 
   Extra Example 1: Always, Sometimes, or Never 12:22 
   Extra Example 2: Always, Sometimes, or Never 13:15 
   Extra Example 3: Always, Sometimes, or Never 14:16 
   Extra Example 4: Always, Sometimes, or Never 15:03 
  Deductive Reasoning 36:03
   Intro 0:00 
   Deductive Reasoning 0:06 
    Definition of Deductive Reasoning 0:07 
   Inductive vs. Deductive 2:51 
    Inductive Reasoning 2:52 
    Deductive reasoning 3:19 
   Law of Detachment 3:47 
    Law of Detachment 3:48 
    Examples of Law of Detachment 4:31 
   Law of Syllogism 7:32 
    Law of Syllogism 7:33 
    Example 1: Making a Conclusion 9:02 
    Example 2: Making a Conclusion 12:54 
   Using Laws of Logic 14:12 
    Example 1: Determine the Logic 14:42 
    Example 2: Determine the Logic 17:02 
   Using Laws of Logic, cont. 18:47 
    Example 3: Determine the Logic 19:03 
    Example 4: Determine the Logic 20:56 
   Extra Example 1: Determine the Conclusion and Law 22:12 
   Extra Example 2: Determine the Conclusion and Law 25:39 
   Extra Example 3: Determine the Logic and Law 29:50 
   Extra Example 4: Determine the Logic and Law 31:27 
  Proofs in Algebra: Properties of Equality 44:31
   Intro 0:00 
   Properties of Equality 0:10 
    Addition Property of Equality 0:28 
    Subtraction Property of Equality 1:10 
    Multiplication Property of Equality 1:41 
    Division Property of Equality 1:55 
    Addition Property of Equality Using Angles 2:46 
   Properties of Equality, cont. 4:10 
    Reflexive Property of Equality 4:11 
    Symmetric Property of Equality 5:24 
    Transitive Property of Equality 6:10 
   Properties of Equality, cont. 7:04 
    Substitution Property of Equality 7:05 
    Distributive Property of Equality 8:34 
   Two Column Proof 9:40 
    Example: Two Column Proof 9:46 
   Proof Example 1 16:13 
   Proof Example 2 23:49 
   Proof Example 3 30:33 
   Extra Example 1: Name the Property of Equality 38:07 
   Extra Example 2: Name the Property of Equality 40:16 
   Extra Example 3: Name the Property of Equality 41:35 
   Extra Example 4: Name the Property of Equality 43:02 
  Proving Segment Relationship 41:02
   Intro 0:00 
   Good Proofs 0:12 
    Five Essential Parts 0:13 
   Proof Reasons 1:38 
    Undefined 1:40 
    Definitions 2:06 
    Postulates 2:42 
    Previously Proven Theorems 3:24 
   Congruence of Segments 4:10 
    Theorem: Congruence of Segments 4:12 
   Proof Example 10:16 
    Proof: Congruence of Segments 10:17 
   Setting Up Proofs 19:13 
    Example: Two Segments with Equal Measures 19:15 
   Setting Up Proofs 21:48 
    Example: Vertical Angles are Congruent 21:50 
   Setting Up Proofs 23:59 
    Example: Segment of a Triangle 24:00 
   Extra Example 1: Congruence of Segments 27:03 
   Extra Example 2: Setting Up Proofs 28:50 
   Extra Example 3: Setting Up Proofs 30:55 
   Extra Example 4: Two-Column Proof 33:11 
  Proving Angle Relationships 33:37
   Intro 0:00 
   Supplement Theorem 0:05 
    Supplementary Angles 0:06 
   Congruence of Angles 2:37 
    Proof: Congruence of Angles 2:38 
   Angle Theorems 6:54 
    Angle Theorem 1: Supplementary Angles 6:55 
    Angle Theorem 2: Complementary Angles 10:25 
   Angle Theorems 11:32 
    Angle Theorem 3: Right Angles 11:35 
    Angle Theorem 4: Vertical Angles 12:09 
    Angle Theorem 5: Perpendicular Lines 12:57 
   Using Angle Theorems 13:45 
    Example 1: Always, Sometimes, or Never 13:50 
    Example 2: Always, Sometimes, or Never 14:28 
    Example 3: Always, Sometimes, or Never 16:21 
   Extra Example 1: Always, Sometimes, or Never 16:53 
   Extra Example 2: Find the Measure of Each Angle 18:55 
   Extra Example 3: Find the Measure of Each Angle 25:03 
   Extra Example 4: Two-Column Proof 27:08 
III. Perpendicular & Parallel Lines
  Parallel Lines and Transversals 37:35
   Intro 0:00 
   Lines 0:06 
    Parallel Lines 0:09 
    Skew Lines 2:02 
    Transversal 3:42 
   Angles Formed by a Transversal 4:28 
    Interior Angles 5:53 
    Exterior Angles 6:09 
    Consecutive Interior Angles 7:04 
    Alternate Exterior Angles 9:47 
    Alternate Interior Angles 11:22 
    Corresponding Angles 12:27 
   Angles Formed by a Transversal 15:29 
    Relationship Between Angles 15:30 
   Extra Example 1: Intersecting, Parallel, or Skew 19:26 
   Extra Example 2: Draw a Diagram 21:37 
   Extra Example 3: Name the Figures 24:12 
   Extra Example 4: Angles Formed by a Transversal 28:38 
  Angles and Parallel Lines 41:53
   Intro 0:00 
   Corresponding Angles Postulate 0:05 
    Corresponding Angles Postulate 0:06 
   Alternate Interior Angles Theorem 3:05 
    Alternate Interior Angles Theorem 3:07 
   Consecutive Interior Angles Theorem 5:16 
    Consecutive Interior Angles Theorem 5:17 
   Alternate Exterior Angles Theorem 6:42 
    Alternate Exterior Angles Theorem 6:43 
   Parallel Lines Cut by a Transversal 7:18 
    Example: Parallel Lines Cut by a Transversal 7:19 
   Perpendicular Transversal Theorem 14:54 
    Perpendicular Transversal Theorem 14:55 
   Extra Example 1: State the Postulate or Theorem 16:37 
   Extra Example 2: Find the Measure of the Numbered Angle 18:53 
   Extra Example 3: Find the Measure of Each Angle 25:13 
   Extra Example 4: Find the Values of x, y, and z 36:26 
  Slope of Lines 44:06
   Intro 0:00 
   Definition of Slope 0:06 
    Slope Equation 0:13 
   Slope of a Line 3:45 
    Example: Find the Slope of a Line 3:47 
   Slope of a Line 8:38 
    More Example: Find the Slope of a Line 8:40 
   Slope Postulates 12:32 
    Proving Slope Postulates 12:33 
   Parallel or Perpendicular Lines 17:23 
    Example: Parallel or Perpendicular Lines 17:24 
   Using Slope Formula 20:02 
    Example: Using Slope Formula 20:03 
   Extra Example 1: Slope of a Line 25:10 
   Extra Example 2: Slope of a Line 26:31 
   Extra Example 3: Graph the Line 34:11 
   Extra Example 4: Using the Slope Formula 38:50 
  Proving Lines Parallel 25:55
   Intro 0:00 
   Postulates 0:06 
    Postulate 1: Parallel Lines 0:21 
    Postulate 2: Parallel Lines 2:16 
   Parallel Postulate 3:28 
    Definition and Example of Parallel Postulate 3:29 
   Theorems 4:29 
    Theorem 1: Parallel Lines 4:40 
    Theorem 2: Parallel Lines 5:37 
   Theorems, cont. 6:10 
    Theorem 3: Parallel Lines 6:11 
   Extra Example 1: Determine Parallel Lines 6:56 
   Extra Example 2: Find the Value of x 11:42 
   Extra Example 3: Opposite Sides are Parallel 14:48 
   Extra Example 4: Proving Parallel Lines 20:42 
  Parallels and Distance 19:48
   Intro 0:00 
   Distance Between a Points and Line 0:07 
    Definition and Example 0:08 
   Distance Between Parallel Lines 1:51 
    Definition and Example 1:52 
   Extra Example 1: Drawing a Segment to Represent Distance 3:02 
   Extra Example 2: Drawing a Segment to Represent Distance 4:27 
   Extra Example 3: Graph, Plot, and Construct a Perpendicular Segment 5:13 
   Extra Example 4: Distance Between Two Parallel Lines 15:37 
IV. Congruent Triangles
  Classifying Triangles 28:43
   Intro 0:00 
   Triangles 0:09 
    Triangle: A Three-Sided Polygon 0:10 
    Sides 1:00 
    Vertices 1:22 
    Angles 1:56 
   Classifying Triangles by Angles 2:59 
    Acute Triangle 3:19 
    Obtuse Triangle 4:08 
    Right Triangle 4:44 
   Equiangular Triangle 5:38 
    Definition and Example of an Equiangular Triangle 5:39 
   Classifying Triangles by Sides 6:57 
    Scalene Triangle 7:17 
    Isosceles Triangle 7:57 
    Equilateral Triangle 8:12 
   Isosceles Triangle 8:58 
    Labeling Isosceles Triangle 9:00 
    Labeling Right Triangle 10:44 
   Isosceles Triangle 11:10 
    Example: Find x, AB, BC, and AC 11:11 
   Extra Example 1: Classify Each Triangle 13:45 
   Extra Example 2: Always, Sometimes, or Never 16:28 
   Extra Example 3: Find All the Sides of the Isosceles Triangle 20:29 
   Extra Example 4: Distance Formula and Triangle 22:29 
  Measuring Angles in Triangles 44:43
   Intro 0:00 
   Angle Sum Theorem 0:09 
    Angle Sum Theorem for Triangle 0:11 
   Using Angle Sum Theorem 4:06 
    Find the Measure of the Missing Angle 4:07 
   Third Angle Theorem 4:58 
    Example: Third Angle Theorem 4:59 
   Exterior Angle Theorem 7:58 
    Example: Exterior Angle Theorem 8:00 
   Flow Proof of Exterior Angle Theorem 15:14 
    Flow Proof of Exterior Angle Theorem 15:17 
   Triangle Corollaries 27:21 
    Triangle Corollary 1 27:50 
    Triangle Corollary 2 30:42 
   Extra Example 1: Find the Value of x 32:55 
   Extra Example 2: Find the Value of x 34:20 
   Extra Example 3: Find the Measure of the Angle 35:38 
   Extra Example 4: Find the Measure of Each Numbered Angle 39:00 
  Exploring Congruent Triangles 26:46
   Intro 0:00 
   Congruent Triangles 0:15 
    Example of Congruent Triangles 0:17 
   Corresponding Parts 3:39 
    Corresponding Angles and Sides of Triangles 3:40 
   Definition of Congruent Triangles 11:24 
    Definition of Congruent Triangles 11:25 
   Triangle Congruence 16:37 
    Congruence of Triangles 16:38 
   Extra Example 1: Congruence Statement 18:24 
   Extra Example 2: Congruence Statement 21:26 
   Extra Example 3: Draw and Label the Figure 23:09 
   Extra Example 4: Drawing Triangles 24:04 
  Proving Triangles Congruent 47:51
   Intro 0:00 
   SSS Postulate 0:18 
    Side-Side-Side Postulate 0:27 
   SAS Postulate 2:26 
    Side-Angle-Side Postulate 2:29 
   SAS Postulate 3:57 
    Proof Example 3:58 
   ASA Postulate 11:47 
    Angle-Side-Angle Postulate 11:53 
   AAS Theorem 14:13 
    Angle-Angle-Side Theorem 14:14 
   Methods Overview 16:16 
    Methods Overview 16:17 
    SSS 16:33 
    SAS 17:06 
    ASA 17:50 
    AAS 18:17 
    CPCTC 19:14 
   Extra Example 1:Proving Triangles are Congruent 21:29 
   Extra Example 2: Proof 25:40 
   Extra Example 3: Proof 30:41 
   Extra Example 4: Proof 38:41 
  Isosceles and Equilateral Triangles 27:53
   Intro 0:00 
   Isosceles Triangle Theorem 0:07 
    Isosceles Triangle Theorem 0:09 
   Isosceles Triangle Theorem 2:26 
    Example: Using the Isosceles Triangle Theorem 2:27 
   Isosceles Triangle Theorem Converse 3:29 
    Isosceles Triangle Theorem Converse 3:30 
   Equilateral Triangle Theorem Corollaries 4:30 
    Equilateral Triangle Theorem Corollary 1 4:59 
    Equilateral Triangle Theorem Corollary 2 5:55 
   Extra Example 1: Find the Value of x 7:08 
   Extra Example 2: Find the Value of x 10:04 
   Extra Example 3: Proof 14:04 
   Extra Example 4: Proof 22:41 
V. Triangle Inequalities
  Special Segments in Triangles 43:44
   Intro 0:00 
   Perpendicular Bisector 0:06 
    Perpendicular Bisector 0:07 
   Perpendicular Bisector 4:07 
    Perpendicular Bisector Theorems 4:08 
   Median 6:30 
    Definition of Median 6:31 
   Median 9:41 
    Example: Median 9:42 
   Altitude 12:22 
    Definition of Altitude 12:23 
   Angle Bisector 14:33 
    Definition of Angle Bisector 14:34 
   Angle Bisector 16:41 
    Angle Bisector Theorems 16:42 
   Special Segments Overview 18:57 
    Perpendicular Bisector 19:04 
    Median 19:32 
    Altitude 19:49 
    Angle Bisector 20:02 
    Examples: Special Segments 20:18 
   Extra Example 1: Draw and Label 22:36 
   Extra Example 2: Draw the Altitudes for Each Triangle 24:37 
   Extra Example 3: Perpendicular Bisector 27:57 
   Extra Example 4: Draw, Label, and Write Proof 34:33 
  Right Triangles 26:34
   Intro 0:00 
   LL Theorem 0:21 
    Leg-Leg Theorem 0:25 
   HA Theorem 2:23 
    Hypotenuse-Angle Theorem 2:24 
   LA Theorem 4:49 
    Leg-Angle Theorem 4:50 
   LA Theorem 6:18 
    Example: Find x and y 6:19 
   HL Postulate 8:22 
    Hypotenuse-Leg Postulate 8:23 
   Extra Example 1: LA Theorem & HL Postulate 10:57 
   Extra Example 2: Find x So That Each Pair of Triangles is Congruent 14:15 
   Extra Example 3: Two-column Proof 17:02 
   Extra Example 4: Two-column Proof 21:01 
  Indirect Proofs and Inequalities 33:30
   Intro 0:00 
   Writing an Indirect Proof 0:09 
    Step 1 0:49 
    Step 2 2:32 
    Step 3 3:00 
   Indirect Proof 4:30 
    Example: 2 + 6 = 8 5:00 
    Example: The Suspect is Guilty 5:40 
    Example: Measure of Angle A < Measure of Angle B 6:06 
   Definition of Inequality 7:47 
    Definition of Inequality & Example 7:48 
   Properties of Inequality 9:55 
    Comparison Property 9:58 
    Transitive Property 10:33 
    Addition and Subtraction Properties 12:01 
    Multiplication and Division Properties 13:07 
   Exterior Angle Inequality Theorem 14:12 
    Example: Exterior Angle Inequality Theorem 14:13 
   Extra Example 1: Draw a Diagram for the Statement 18:32 
   Extra Example 2: Name the Property for Each Statement 19:56 
   Extra Example 3: State the Assumption 21:22 
   Extra Example 4: Write an Indirect Proof 25:39 
  Inequalities for Sides and Angles of a Triangle 17:26
   Intro 0:00 
   Side to Angles 0:10 
    If One Side of a Triangle is Longer Than Another Side 0:11 
   Converse: Angles to Sides 1:57 
    If One Angle of a Triangle Has a Greater Measure Than Another Angle 1:58 
   Extra Example 1: Name the Angles in the Triangle From Least to Greatest 2:38 
   Extra Example 2: Find the Longest and Shortest Segment in the Triangle 3:47 
   Extra Example 3: Angles and Sides of a Triangle 4:51 
   Extra Example 4: Two-column Proof 9:08 
  Triangle Inequality 28:11
   Intro 0:00 
   Triangle Inequality Theorem 0:05 
    Triangle Inequality Theorem 0:06 
   Triangle Inequality Theorem 4:22 
    Example 1: Triangle Inequality Theorem 4:23 
    Example 2: Triangle Inequality Theorem 9:40 
   Extra Example 1: Determine if the Three Numbers can Represent the Sides of a Triangle 12:00 
   Extra Example 2: Finding the Third Side of a Triangle 13:34 
   Extra Example 3: Always True, Sometimes True, or Never True 18:18 
   Extra Example 4: Triangle and Vertices 22:36 
  Inequalities Involving Two Triangles 29:36
   Intro 0:00 
   SAS Inequality Theorem 0:06 
    SAS Inequality Theorem & Example 0:25 
   SSS Inequality Theorem 4:33 
    SSS Inequality Theorem & Example 4:34 
   Extra Example 1: Write an Inequality Comparing the Segments 6:08 
   Extra Example 2: Determine if the Statement is True 9:52 
   Extra Example 3: Write an Inequality for x 14:20 
   Extra Example 4: Two-column Proof 17:44 
VI. Quadrilaterals
  Parallelograms 29:11
   Intro 0:00 
   Quadrilaterals 0:06 
    Four-sided Polygons 0:08 
    Non Examples of Quadrilaterals 0:47 
   Parallelograms 1:35 
    Parallelograms 1:36 
   Properties of Parallelograms 4:28 
    Opposite Sides of a Parallelogram are Congruent 4:29 
    Opposite Angles of a Parallelogram are Congruent 5:49 
   Angles and Diagonals 6:24 
    Consecutive Angles in a Parallelogram are Supplementary 6:25 
    The Diagonals of a Parallelogram Bisect Each Other 8:42 
   Extra Example 1: Complete Each Statement About the Parallelogram 10:26 
   Extra Example 2: Find the Values of x, y, and z of the Parallelogram 13:21 
   Extra Example 3: Find the Distance of Each Side to Verify the Parallelogram 16:35 
   Extra Example 4: Slope of Parallelogram 23:15 
  Proving Parallelograms 42:43
   Intro 0:00 
   Parallelogram Theorems 0:09 
    Theorem 1 0:20 
    Theorem 2 1:50 
   Parallelogram Theorems, Cont. 3:10 
    Theorem 3 3:11 
    Theorem 4 4:15 
   Proving Parallelogram 6:21 
    Example: Determine if Quadrilateral ABCD is a Parallelogram 6:22 
   Summary 14:01 
    Both Pairs of Opposite Sides are Parallel 14:14 
    Both Pairs of Opposite Sides are Congruent 15:09 
    Both Pairs of Opposite Angles are Congruent 15:24 
    Diagonals Bisect Each Other 15:44 
    A Pair of Opposite Sides is Both Parallel and Congruent 16:13 
   Extra Example 1: Determine if Each Quadrilateral is a Parallelogram 16:54 
   Extra Example 2: Find the Value of x and y 20:23 
   Extra Example 3: Determine if the Quadrilateral ABCD is a Parallelogram 24:05 
   Extra Example 4: Two-column Proof 30:28 
  Rectangles 29:47
   Intro 0:00 
   Rectangles 0:03 
    Definition of Rectangles 0:04 
   Diagonals of Rectangles 2:52 
    Rectangles: Diagonals Property 1 2:53 
    Rectangles: Diagonals Property 2 3:30 
   Proving a Rectangle 4:40 
    Example: Determine Whether Parallelogram ABCD is a Rectangle 4:41 
   Rectangles Summary 9:22 
    Opposite Sides are Congruent and Parallel 9:40 
    Opposite Angles are Congruent 9:51 
    Consecutive Angles are Supplementary 9:58 
    Diagonals are Congruent and Bisect Each Other 10:05 
    All Four Angles are Right Angles 10:40 
   Extra Example 1: Find the Value of x 11:03 
   Extra Example 2: Name All Congruent Sides and Angles 13:52 
   Extra Example 3: Always, Sometimes, or Never True 19:39 
   Extra Example 4: Determine if ABCD is a Rectangle 26:45 
  Squares and Rhombi 39:14
   Intro 0:00 
   Rhombus 0:09 
    Definition of a Rhombus 0:10 
   Diagonals of a Rhombus 2:03 
    Rhombus: Diagonals Property 1 2:21 
    Rhombus: Diagonals Property 2 3:49 
    Rhombus: Diagonals Property 3 4:36 
   Rhombus 6:17 
    Example: Use the Rhombus to Find the Missing Value 6:18 
   Square 8:17 
    Definition of a Square 8:20 
   Summary Chart 11:06 
    Parallelogram 11:07 
    Rectangle 12:56 
    Rhombus 13:54 
    Square 14:44 
   Extra Example 1: Diagonal Property 15:44 
   Extra Example 2: Use Rhombus ABCD to Find the Missing Value 19:39 
   Extra Example 3: Always, Sometimes, or Never True 23:06 
   Extra Example 4: Determine the Quadrilateral 28:02 
  Trapezoids and Kites 30:48
   Intro 0:00 
   Trapezoid 0:10 
    Definition of Trapezoid 0:12 
   Isosceles Trapezoid 2:57 
    Base Angles of an Isosceles Trapezoid 2:58 
    Diagonals of an Isosceles Trapezoid 4:05 
   Median of a Trapezoid 4:26 
    Median of a Trapezoid 4:27 
   Median of a Trapezoid 6:41 
    Median Formula 7:00 
   Kite 8:28 
    Definition of a Kite 8:29 
   Quadrilaterals Summary 11:19 
    A Quadrilateral with Two Pairs of Adjacent Congruent Sides 11:20 
   Extra Example 1: Isosceles Trapezoid 14:50 
   Extra Example 2: Median of Trapezoid 18:28 
   Extra Example 3: Always, Sometimes, or Never 24:13 
   Extra Example 4: Determine if the Figure is a Trapezoid 26:49 
VII. Proportions and Similarity
  Using Proportions and Ratios 20:10
   Intro 0:00 
   Ratio 0:05 
    Definition and Examples of Writing Ratio 0:06 
   Proportion 2:05 
    Definition of Proportion 2:06 
    Examples of Proportion 2:29 
   Using Ratio 5:53 
    Example: Ratio 5:54 
   Extra Example 1: Find Three Ratios Equivalent to 2/5 9:28 
   Extra Example 2: Proportion and Cross Products 10:32 
   Extra Example 3: Express Each Ratio as a Fraction 13:18 
   Extra Example 4: Fin the Measure of a 3:4:5 Triangle 17:26 
  Similar Polygons 27:53
   Intro 0:00 
   Similar Polygons 0:05 
    Definition of Similar Polygons 0:06 
    Example of Similar Polygons 2:32 
   Scale Factor 4:26 
    Scale Factor: Definition and Example 4:27 
   Extra Example 1: Determine if Each Pair of Figures is Similar 7:03 
   Extra Example 2: Find the Values of x and y 11:33 
   Extra Example 3: Similar Triangles 19:57 
   Extra Example 4: Draw Two Similar Figures 23:36 
  Similar Triangles 34:10
   Intro 0:00 
   AA Similarity 0:10 
    Definition of AA Similarity 0:20 
    Example of AA Similarity 2:32 
   SSS Similarity 4:46 
    Definition of SSS Similarity 4:47 
    Example of SSS Similarity 6:00 
   SAS Similarity 8:04 
    Definition of SAS Similarity 8:05 
    Example of SAS Similarity 9:12 
   Extra Example 1: Determine Whether Each Pair of Triangles is Similar 10:59 
   Extra Example 2: Determine Which Triangles are Similar 16:08 
   Extra Example 3: Determine if the Statement is True or False 23:11 
   Extra Example 4: Write Two-Column Proof 26:25 
  Parallel Lines and Proportional Parts 24:07
   Intro 0:00 
   Triangle Proportionality 0:07 
    Definition of Triangle Proportionality 0:08 
    Example of Triangle Proportionality 0:51 
   Triangle Proportionality Converse 2:19 
    Triangle Proportionality Converse 2:20 
   Triangle Mid-segment 3:42 
    Triangle Mid-segment: Definition and Example 3:43 
   Parallel Lines and Transversal 6:51 
    Parallel Lines and Transversal 6:52 
   Extra Example 1: Complete Each Statement 8:59 
   Extra Example 2: Determine if the Statement is True or False 12:28 
   Extra Example 3: Find the Value of x and y 15:35 
   Extra Example 4: Find Midpoints of a Triangle 20:43 
  Parts of Similar Triangles 27:06
   Intro 0:00 
   Proportional Perimeters 0:09 
    Proportional Perimeters: Definition and Example 0:10 
   Similar Altitudes 2:23 
    Similar Altitudes: Definition and Example 2:24 
   Similar Angle Bisectors 4:50 
    Similar Angle Bisectors: Definition and Example 4:51 
   Similar Medians 6:05 
    Similar Medians: Definition and Example 6:06 
   Angle Bisector Theorem 7:33 
    Angle Bisector Theorem 7:34 
   Extra Example 1: Parts of Similar Triangles 10:52 
   Extra Example 2: Parts of Similar Triangles 14:57 
   Extra Example 3: Parts of Similar Triangles 19:27 
   Extra Example 4: Find the Perimeter of Triangle ABC 23:14 
VIII. Applying Right Triangles & Trigonometry
  Pythagorean Theorem 21:14
   Intro 0:00 
   Pythagorean Theorem 0:05 
    Pythagorean Theorem & Example 0:06 
   Pythagorean Converse 1:20 
    Pythagorean Converse & Example 1:21 
   Pythagorean Triple 2:42 
    Pythagorean Triple 2:43 
   Extra Example 1: Find the Missing Side 4:59 
   Extra Example 2: Determine Right Triangle 7:40 
   Extra Example 3: Determine Pythagorean Triple 11:30 
   Extra Example 4: Vertices and Right Triangle 14:29 
  Geometric Mean 40:59
   Intro 0:00 
   Geometric Mean 0:04 
    Geometric Mean & Example 0:05 
   Similar Triangles 4:32 
    Similar Triangles 4:33 
   Geometric Mean-Altitude 11:10 
    Geometric Mean-Altitude & Example 11:11 
   Geometric Mean-Leg 14:47 
    Geometric Mean-Leg & Example 14:18 
   Extra Example 1: Geometric Mean Between Each Pair of Numbers 20:10 
   Extra Example 2: Similar Triangles 23:46 
   Extra Example 3: Geometric Mean of Triangles 28:30 
   Extra Example 4: Geometric Mean of Triangles 36:58 
  Special Right Triangles 37:57
   Intro 0:00 
   45-45-90 Triangles 0:06 
    Definition of 45-45-90 Triangles 0:25 
   45-45-90 Triangles 5:51 
    Example: Find n 5:52 
   30-60-90 Triangles 8:59 
    Definition of 30-60-90 Triangles 9:00 
   30-60-90 Triangles 12:25 
    Example: Find n 12:26 
   Extra Example 1: Special Right Triangles 15:08 
   Extra Example 2: Special Right Triangles 18:22 
   Extra Example 3: Word Problems & Special Triangles 27:40 
   Extra Example 4: Hexagon & Special Triangles 33:51 
  Ratios in Right Triangles 40:37
   Intro 0:00 
   Trigonometric Ratios 0:08 
    Definition of Trigonometry 0:13 
    Sine (sin), Cosine (cos), & Tangent (tan) 0:50 
   Trigonometric Ratios 3:04 
    Trig Functions 3:05 
    Inverse Trig Functions 5:02 
   SOHCAHTOA 8:16 
    sin x 9:07 
    cos x 10:00 
    tan x 10:32 
    Example: SOHCAHTOA & Triangle 12:10 
   Extra Example 1: Find the Value of Each Ratio or Angle Measure 14:36 
   Extra Example 2: Find Sin, Cos, and Tan 18:51 
   Extra Example 3: Find the Value of x Using SOHCAHTOA 22:55 
   Extra Example 4: Trigonometric Ratios in Right Triangles 32:13 
  Angles of Elevation and Depression 21:04
   Intro 0:00 
   Angle of Elevation 0:10 
    Definition of Angle of Elevation & Example 0:11 
   Angle of Depression 1:19 
    Definition of Angle of Depression & Example 1:20 
   Extra Example 1: Name the Angle of Elevation and Depression 2:22 
   Extra Example 2: Word Problem & Angle of Depression 4:41 
   Extra Example 3: Word Problem & Angle of Elevation 14:02 
   Extra Example 4: Find the Missing Measure 18:10 
  Law of Sines 35:25
   Intro 0:00 
   Law of Sines 0:20 
    Law of Sines 0:21 
   Law of Sines 3:34 
    Example: Find b 3:35 
   Solving the Triangle 9:19 
    Example: Using the Law of Sines to Solve Triangle 9:20 
   Extra Example 1: Law of Sines and Triangle 17:43 
   Extra Example 2: Law of Sines and Triangle 20:06 
   Extra Example 3: Law of Sines and Triangle 23:54 
   Extra Example 4: Law of Sines and Triangle 28:59 
  Law of Cosines 52:43
   Intro 0:00 
   Law of Cosines 0:35 
    Law of Cosines 0:36 
   Law of Cosines 6:22 
    Use the Law of Cosines When Both are True 6:23 
   Law of Cosines 8:35 
    Example: Law of Cosines 8:36 
   Extra Example 1: Law of Sines or Law of Cosines? 13:35 
   Extra Example 2: Use the Law of Cosines to Find the Missing Measure 17:02 
   Extra Example 3: Solve the Triangle 30:49 
   Extra Example 4: Find the Measure of Each Diagonal of the Parallelogram 41:39 
IX. Circles
  Segments in a Circle 22:43
   Intro 0:00 
   Segments in a Circle 0:10 
    Circle 0:11 
    Chord 0:59 
    Diameter 1:32 
    Radius 2:07 
    Secant 2:17 
    Tangent 3:10 
   Circumference 3:56 
    Introduction to Circumference 3:57 
    Example: Find the Circumference of the Circle 5:09 
   Circumference 6:40 
    Example: Find the Circumference of the Circle 6:41 
   Extra Example 1: Use the Circle to Answer the Following 9:10 
   Extra Example 2: Find the Missing Measure 12:53 
   Extra Example 3: Given the Circumference, Find the Perimeter of the Triangle 15:51 
   Extra Example 4: Find the Circumference of Each Circle 19:24 
  Angles and Arc 35:24
   Intro 0:00 
   Central Angle 0:06 
    Definition of Central Angle 0:07 
   Sum of Central Angles 1:17 
    Sum of Central Angles 1:18 
   Arcs 2:27 
    Minor Arc 2:30 
    Major Arc 3:47 
   Arc Measure 5:24 
    Measure of Minor Arc 5:24 
    Measure of Major Arc 6:53 
    Measure of a Semicircle 7:11 
   Arc Addition Postulate 8:25 
    Arc Addition Postulate 8:26 
   Arc Length 9:43 
    Arc Length and Example 9:44 
   Concentric Circles 16:05 
    Concentric Circles 16:06 
   Congruent Circles and Arcs 17:50 
    Congruent Circles 17:51 
    Congruent Arcs 18:47 
   Extra Example 1: Minor Arc, Major Arc, and Semicircle 20:14 
   Extra Example 2: Measure and Length of Arc 22:52 
   Extra Example 3: Congruent Arcs 25:48 
   Extra Example 4: Angles and Arcs 30:33 
  Arcs and Chords 21:51
   Intro 0:00 
   Arcs and Chords 0:07 
    Arc of the Chord 0:08 
    Theorem 1: Congruent Minor Arcs 1:01 
   Inscribed Polygon 2:10 
    Inscribed Polygon 2:11 
   Arcs and Chords 3:18 
    Theorem 2: When a Diameter is Perpendicular to a Chord 3:19 
   Arcs and Chords 5:05 
    Theorem 3: Congruent Chords 5:06 
   Extra Example 1: Congruent Arcs 10:35 
   Extra Example 2: Length of Arc 13:50 
   Extra Example 3: Arcs and Chords 17:09 
   Extra Example 4: Arcs and Chords 19:45 
  Inscribed Angles 27:53
   Intro 0:00 
   Inscribed Angles 0:07 
    Definition of Inscribed Angles 0:08 
   Inscribed Angles 0:58 
    Inscribed Angle Theorem 1 0:59 
   Inscribed Angles 3:29 
    Inscribed Angle Theorem 2 3:30 
   Inscribed Angles 4:38 
    Inscribed Angle Theorem 3 4:39 
   Inscribed Quadrilateral 5:50 
    Inscribed Quadrilateral 5:51 
   Extra Example 1: Central Angle, Inscribed Angle, and Intercepted Arc 7:02 
   Extra Example 2: Inscribed Angles 9:24 
   Extra Example 3: Inscribed Angles 14:00 
   Extra Example 4: Complete the Proof 17:58 
  Tangents 26:16
   Intro 0:00 
   Tangent Theorems 0:04 
    Tangent Theorem 1 0:05 
    Tangent Theorem 1 Converse 0:55 
   Common Tangents 1:34 
    Common External Tangent 2:12 
    Common Internal Tangent 2:30 
   Tangent Segments 3:08 
    Tangent Segments 3:09 
   Circumscribed Polygons 4:11 
    Circumscribed Polygons 4:12 
   Extra Example 1: Tangents & Circumscribed Polygons 5:50 
   Extra Example 2: Tangents & Circumscribed Polygons 8:35 
   Extra Example 3: Tangents & Circumscribed Polygons 11:50 
   Extra Example 4: Tangents & Circumscribed Polygons 15:43 
  Secants, Tangents, & Angle Measures 27:50
   Intro 0:00 
   Secant 0:08 
    Secant 0:09 
   Secant and Tangent 0:49 
    Secant and Tangent 0:50 
   Interior Angles 2:56 
    Secants & Interior Angles 2:57 
   Exterior Angles 7:21 
    Secants & Exterior Angles 7:22 
   Extra Example 1: Secants, Tangents, & Angle Measures 10:53 
   Extra Example 2: Secants, Tangents, & Angle Measures 13:31 
   Extra Example 3: Secants, Tangents, & Angle Measures 19:54 
   Extra Example 4: Secants, Tangents, & Angle Measures 22:29 
  Special Segments in a Circle 23:08
   Intro 0:00 
   Chord Segments 0:05 
    Chord Segments 0:06 
   Secant Segments 1:36 
    Secant Segments 1:37 
   Tangent and Secant Segments 4:10 
    Tangent and Secant Segments 4:11 
   Extra Example 1: Special Segments in a Circle 5:53 
   Extra Example 2: Special Segments in a Circle 7:58 
   Extra Example 3: Special Segments in a Circle 11:24 
   Extra Example 4: Special Segments in a Circle 18:09 
  Equations of Circles 27:01
   Intro 0:00 
   Equation of a Circle 0:06 
    Standard Equation of a Circle 0:07 
    Example 1: Equation of a Circle 0:57 
    Example 2: Equation of a Circle 1:36 
   Extra Example 1: Determine the Coordinates of the Center and the Radius 4:56 
   Extra Example 2: Write an Equation Based on the Given Information 7:53 
   Extra Example 3: Graph Each Circle 16:48 
   Extra Example 4: Write the Equation of Each Circle 19:17 
X. Polygons & Area
  Polygons 27:24
   Intro 0:00 
   Polygons 0:10 
    Polygon vs. Not Polygon 0:18 
   Convex and Concave 1:46 
    Convex vs. Concave Polygon 1:52 
   Regular Polygon 4:04 
    Regular Polygon 4:05 
   Interior Angle Sum Theorem 4:53 
    Triangle 5:03 
    Quadrilateral 6:05 
    Pentagon 6:38 
    Hexagon 7:59 
    20-Gon 9:36 
   Exterior Angle Sum Theorem 12:04 
    Exterior Angle Sum Theorem 12:05 
   Extra Example 1: Drawing Polygons 13:51 
   Extra Example 2: Convex Polygon 15:16 
   Extra Example 3: Exterior Angle Sum Theorem 18:21 
   Extra Example 4: Interior Angle Sum Theorem 22:20 
  Area of Parallelograms 17:46
   Intro 0:00 
   Parallelograms 0:06 
    Definition and Area Formula 0:07 
   Area of Figure 2:00 
    Area of Figure 2:01 
   Extra Example 1:Find the Area of the Shaded Area 3:14 
   Extra Example 2: Find the Height and Area of the Parallelogram 6:00 
   Extra Example 3: Find the Area of the Parallelogram Given Coordinates and Vertices 10:11 
   Extra Example 4: Find the Area of the Figure 14:31 
  Area of Triangles Rhombi, & Trapezoids 20:31
   Intro 0:00 
   Area of a Triangle 0:06 
    Area of a Triangle: Formula and Example 0:07 
   Area of a Trapezoid 2:31 
    Area of a Trapezoid: Formula 2:32 
    Area of a Trapezoid: Example 6:55 
   Area of a Rhombus 8:05 
    Area of a Rhombus: Formula and Example 8:06 
   Extra Example 1: Find the Area of the Polygon 9:51 
   Extra Example 2: Find the Area of the Figure 11:19 
   Extra Example 3: Find the Area of the Figure 14:16 
   Extra Example 4: Find the Height of the Trapezoid 18:10 
  Area of Regular Polygons & Circles 36:43
   Intro 0:00 
   Regular Polygon 0:08 
    SOHCAHTOA 0:54 
    30-60-90 Triangle 1:52 
    45-45-90 Triangle 2:40 
   Area of a Regular Polygon 3:39 
    Area of a Regular Polygon 3:40 
   Are of a Circle 7:55 
    Are of a Circle 7:56 
   Extra Example 1: Find the Area of the Regular Polygon 8:22 
   Extra Example 2: Find the Area of the Regular Polygon 16:48 
   Extra Example 3: Find the Area of the Shaded Region 24:11 
   Extra Example 4: Find the Area of the Shaded Region 32:24 
  Perimeter & Area of Similar Figures 18:17
   Intro 0:00 
   Perimeter of Similar Figures 0:08 
    Example: Scale Factor & Perimeter of Similar Figures 0:09 
   Area of Similar Figures 2:44 
    Example:Scale Factor & Area of Similar Figures 2:55 
   Extra Example 1: Complete the Table 6:09 
   Extra Example 2: Find the Ratios of the Perimeter and Area of the Similar Figures 8:56 
   Extra Example 3: Find the Unknown Area 12:04 
   Extra Example 4: Use the Given Area to Find AB 14:26 
  Geometric Probability 38:40
   Intro 0:00 
   Length Probability Postulate 0:05 
    Length Probability Postulate 0:06 
   Are Probability Postulate 2:34 
    Are Probability Postulate 2:35 
   Are of a Sector of a Circle 4:11 
    Are of a Sector of a Circle Formula 4:12 
    Are of a Sector of a Circle Example 7:51 
   Extra Example 1: Length Probability 11:07 
   Extra Example 2: Area Probability 12:14 
   Extra Example 3: Area Probability 17:17 
   Extra Example 4: Area of a Sector of a Circle 26:23 
XI. Solids
  Three-Dimensional Figures 23:39
   Intro 0:00 
   Polyhedrons 0:05 
    Polyhedrons: Definition and Examples 0:06 
    Faces 1:08 
    Edges 1:55 
    Vertices 2:23 
   Solids 2:51 
    Pyramid 2:54 
    Cylinder 3:45 
    Cone 4:09 
    Sphere 4:23 
   Prisms 5:00 
     Rectangular, Regular, and Cube Prisms 5:02 
   Platonic Solids 9:48 
    Five Types of Regular Polyhedra 9:49 
   Slices and Cross Sections 12:07 
    Slices 12:08 
    Cross Sections 12:47 
   Extra Example 1: Name the Edges, Faces, and Vertices of the Polyhedron 14:23 
   Extra Example 2: Determine if the Figure is a Polyhedron and Explain Why 17:37 
   Extra Example 3: Describe the Slice Resulting from the Cut 19:12 
   Extra Example 4: Describe the Shape of the Intersection 21:25 
  Surface Area of Prisms and Cylinders 38:50
   Intro 0:00 
   Prisms 0:06 
    Bases 0:07 
    Lateral Faces 0:52 
    Lateral Edges 1:19 
    Altitude 1:58 
   Prisms 2:24 
    Right Prism 2:25 
    Oblique Prism 2:56 
   Classifying Prisms 3:27 
    Right Rectangular Prism 3:28 
     4:55 
    Oblique Pentagonal Prism 6:26 
    Right Hexagonal Prism 7:14 
   Lateral Area of a Prism 7:42 
    Lateral Area of a Prism 7:43 
   Surface Area of a Prism 13:44 
    Surface Area of a Prism 13:45 
   Cylinder 16:18 
    Cylinder: Right and Oblique 16:19 
   Lateral Area of a Cylinder 18:02 
    Lateral Area of a Cylinder 18:03 
   Surface Area of a Cylinder 20:54 
    Surface Area of a Cylinder 20:55 
   Extra Example 1: Find the Lateral Area and Surface Are of the Prism 21:51 
   Extra Example 2: Find the Lateral Area of the Prism 28:15 
   Extra Example 3: Find the Surface Area of the Prism 31:57 
   Extra Example 4: Find the Lateral Area and Surface Area of the Cylinder 34:17 
  Surface Area of Pyramids and Cones 26:10
   Intro 0:00 
   Pyramids 0:07 
    Pyramids 0:08 
   Regular Pyramids 1:52 
    Regular Pyramids 1:53 
   Lateral Area of a Pyramid 4:33 
    Lateral Area of a Pyramid 4:34 
   Surface Area of a Pyramid 9:19 
    Surface Area of a Pyramid 9:20 
   Cone 10:09 
    Right and Oblique Cone 10:10 
   Lateral Area and Surface Area of a Right Cone 11:20 
    Lateral Area and Surface Are of a Right Cone 11:21 
   Extra Example 1: Pyramid and Prism 13:11 
   Extra Example 2: Find the Lateral Area of the Regular Pyramid 15:00 
   Extra Example 3: Find the Surface Area of the Pyramid 18:29 
   Extra Example 4: Find the Lateral Area and Surface Area of the Cone 22:08 
  Volume of Prisms and Cylinders 21:59
   Intro 0:00 
   Volume of Prism 0:08 
    Volume of Prism 0:10 
   Volume of Cylinder 3:38 
    Volume of Cylinder 3:39 
   Extra Example 1: Find the Volume of the Prism 5:10 
   Extra Example 2: Find the Volume of the Cylinder 8:03 
   Extra Example 3: Find the Volume of the Prism 9:35 
   Extra Example 4: Find the Volume of the Solid 19:06 
  Volume of Pyramids and Cones 22:02
   Intro 0:00 
   Volume of a Cone 0:08 
    Volume of a Cone: Example 0:10 
   Volume of a Pyramid 3:02 
    Volume of a Pyramid: Example 3:03 
   Extra Example 1: Find the Volume of the Pyramid 4:56 
   Extra Example 2: Find the Volume of the Solid 6:01 
   Extra Example 3: Find the Volume of the Pyramid 10:28 
   Extra Example 4: Find the Volume of the Octahedron 16:23 
  Surface Area and Volume of Spheres 14:46
   Intro 0:00 
   Special Segments 0:06 
    Radius 0:07 
    Chord 0:31 
    Diameter 0:55 
    Tangent 1:20 
   Sphere 1:43 
    Plane & Sphere 1:44 
    Hemisphere 2:56 
   Surface Area of a Sphere 3:25 
    Surface Area of a Sphere 3:26 
   Volume of a Sphere 4:08 
    Volume of a Sphere 4:09 
   Extra Example 1: Determine Whether Each Statement is True or False 4:24 
   Extra Example 2: Find the Surface Area of the Sphere 6:17 
   Extra Example 3: Find the Volume of the Sphere with a Diameter of 20 Meters 7:25 
   Extra Example 4: Find the Surface Area and Volume of the Solid 9:17 
  Congruent and Similar Solids 16:06
   Intro 0:00 
   Scale Factor 0:06 
    Scale Factor: Definition and Example 0:08 
   Congruent Solids 1:09 
    Congruent Solids 1:10 
   Similar Solids 2:17 
    Similar Solids 2:18 
   Extra Example 1: Determine if Each Pair of Solids is Similar, Congruent, or Neither 3:35 
   Extra Example 2: Determine if Each Statement is True or False 7:47 
   Extra Example 3: Find the Scale Factor and the Ratio of the Surface Areas and Volume 10:14 
   Extra Example 4: Find the Volume of the Larger Prism 12:14 
XII. Transformational Geometry
  Mapping 14:12
   Intro 0:00 
   Transformation 0:04 
    Rotation 0:32 
    Translation 1:03 
    Reflection 1:17 
    Dilation 1:24 
   Transformations 1:45 
    Examples 1:46 
   Congruence Transformation 2:51 
    Congruence Transformation 2:52 
   Extra Example 1: Describe the Transformation that Occurred in the Mappings 3:37 
   Extra Example 2: Determine if the Transformation is an Isometry 5:16 
   Extra Example 3: Isometry 8:16 
  Reflections 23:17
   Intro 0:00 
   Reflection 0:05 
    Definition of Reflection 0:06 
    Line of Reflection 0:35 
    Point of Reflection 1:22 
   Symmetry 1:59 
    Line of Symmetry 2:00 
    Point of Symmetry 2:48 
   Extra Example 1: Draw the Image over the Line of Reflection and the Point of Reflection 3:45 
   Extra Example 2: Determine Lines and Point of Symmetry 6:59 
   Extra Example 3: Graph the Reflection of the Polygon 11:15 
   Extra Example 4: Graph the Coordinates 16:07 
  Translations 18:43
   Intro 0:00 
   Translation 0:05 
    Translation: Preimage & Image 0:06 
    Example 0:56 
   Composite of Reflections 6:28 
    Composite of Reflections 6:29 
   Extra Example 1: Translation 7:48 
   Extra Example 2: Image, Preimage, and Translation 12:38 
   Extra Example 3: Find the Translation Image Using a Composite of Reflections 15:08 
   Extra Example 4: Find the Value of Each Variable in the Translation 17:18 
  Rotations 21:26
   Intro 0:00 
   Rotations 0:04 
    Rotations 0:05 
   Performing Rotations 2:13 
    Composite of Two Successive Reflections over Two Intersecting Lines 2:14 
    Angle of Rotation: Angle Formed by Intersecting Lines 4:29 
   Angle of Rotation 5:30 
    Rotation Postulate 5:31 
   Extra Example 1: Find the Rotated Image 7:32 
   Extra Example 2: Rotations and Coordinate Plane 10:33 
   Extra Example 3: Find the Value of Each Variable in the Rotation 14:29 
   Extra Example 4: Draw the Polygon Rotated 90 Degree Clockwise about P 16:13 
  Dilation 37:06
   Intro 0:00 
   Dilations 0:06 
    Dilations 0:07 
   Scale Factor 1:36 
    Scale Factor 1:37 
    Example 1 2:06 
    Example 2 6:22 
   Scale Factor 8:20 
    Positive Scale Factor 8:21 
    Negative Scale Factor 9:25 
    Enlargement 12:43 
    Reduction 13:52 
   Extra Example 1: Find the Scale Factor 16:39 
   Extra Example 2: Find the Measure of the Dilation Image 19:32 
   Extra Example 3: Find the Coordinates of the Image with Scale Factor and the Origin as the Center of Dilation 26:18 
   Extra Example 4: Graphing Polygon, Dilation, and Scale Factor 32:08 

Hello; welcome to Educator.com.0000

This is the Geometry course; the very first lesson is on the coordinate plane, which should be somewhat of a review.0002

So, make sure to check out the other free lessons of the syllabus.0008

Let's begin: the coordinate system: the coordinate plane is part of the coordinate system.0011

This here is called the coordinate plane; right here, this is the x-axis, and this is the y-axis.0020

And these two make up the four quadrants of the coordinate plane.0034

If we were to label these, this is I, II, III, and so on; we know that if we go this way, we are going to be going negative; positive and negative.0042

Now, these axes make up four quadrants; the four sections of the coordinate plane are known as quadrants.0063

Here, this is the first quadrant, so this is quadrant I; around this side, we have quadrant II; this is quadrant III, and quadrant IV.0073

So, it starts here, and it goes this way: I, II, III, IV.0085

And for quadrant I, we have a positive; we are only dealing with the positive x-axis and the positive y-axis.0091

For quadrant II, we have a negative x-axis, and then the positive y-axis.0100

For quadrant III, it is negative x and negative y; quadrant IV is positive x and negative y.0104

Those are quadrants; make sure that you remember that there are four of them.0115

The origin is right there: this is known as the origin.0120

The origin is (0,0): the x is 0, and the y is 0--right where they meet, that is the origin.0128

And this is an example of an ordered pair: an ordered pair is when you have the x-coordinate paired with a y-coordinate.0138

And together, it is called an ordered pair.0157

So, if I have a point, (4,2), this would be an ordered pair; my x-coordinate is before, and then 2 would be my y-coordinate.0161

So, let's practice graphing using the coordinate plane: we are going to look for these points and write the coordinates.0177

Here is A, B, and C; for point A, we know that this is 0; this is x; this is y; here is 1, 2, 3, -1, -2, -3;0188

so for point A, we always start with the x-axis first; so the x-coordinate goes first, and that is 1;0211

and then what is my y? 1; that is my ordered pair.0224

For B: my x-coordinate for point B is -1, and my y is going to be -2; so here is (-1,-2).0230

And for C, it is 2 for my x and -1 for my y.0246

We are going to graph each of the points on the coordinate plane.0258

For A, I have (4,2); this is my x and this is my y; this is my x and this is my y.0280

So, I am going to go to positive 4 on this side: 1, 2, 3, 4; and 2 on my y is +2, which is there; (4,2) is going to be right there.0289

I am going to label that point A.0305

For B, (-3,0): x is -3, which is 1, 2, -3...and my y is 0; that means I do not go up or down anything--I stay right there.0309

OK, this is where y is 0; and this part right here is going to be B.0322

C: it will be 1, 2, and 1 right there for C; and then D is (-4,-2); OK.0329

OK, there are all my points on the coordinate plane.0354

One more thing to go over: collinear points are points that lie on the same line.0359

When we have points that line up--you can draw a line through those points--those points will be collinear.0366

And let's see if we have any collinear points here.0377

Well, if you remember from algebra, for slope, we have rise over run;0380

all of those have to do with a line and points, when we are graphing lines on the coordinate plane.0387

So, here we have A and C--we know that those two, or any two, points will be collinear,0393

because you can draw a line through any two points.0401

Here, I know that those three points, A, C, and D, are collinear, because they will be on the same line.0406

They lie on the same line, so A, C, and D are collinear.0419

If you want to double-check this, you can use what you learned from algebra; you can count rise over run.0425

Find your slope from D to C, and then from C to A; and it should be the same, and also from C to A and D to A.0433

OK, so points A, C, and D are collinear.0441

Write the coordinates and quadrants for each point: let's look at point A.0455

Point A: this is -1, and my y is -1, -2, and -3; so for A, -1 is my x-coordinate, and -3 is my y-coordinate.0464

And this is quadrant III, because it goes I, II, and III; so this is in quadrant III.0484

For B: my x is +1, and my y is +2; and that is in quadrant I.0494

C is 3, and my y is -1; and that is quadrant IV; D is -3, and 3; quadrant II.0507

OK, let's do another example: Name two points in each of the four quadrants.0530

OK, we have quadrant I; now quadrant I, I know, is here; quadrant II, quadrant III, and quadrant IV.0536

Quadrant I is going to be positive, and then my y-coordinate is going to be positive.0549

Quadrant II: my x (x is always first), we know, is negative; and then y is positive.0558

Quadrant III is negative for the x-coordinate and negative for my y.0568

Quadrant IV is positive and negative.0575

They should all be different; their signs will be different for each of the quadrants.0578

So, I can name any point; as long as my x-coordinates and my y-coordinates are both positive, they are going to be from Quadrant I.0583

I can just say (1,2) and then maybe (3,4); those are two points from Quadrant I.0592

Quadrant II will be...we have to have a negative x-coordinate and a positive y-coordinate, so what about (-1,2) and (-3,4).0602

Now, you can use your own numbers; you can use the same numbers.0619

As long as you have a negative x and a positive y, they are from Quadrant II.0625

Quadrant III: x and y are both negative, so (-1,-2) and (-3,-4) will be from Quadrant III.0630

And then, Quadrant IV: we have a positive x and a negative y, so (1,2) and (3,-4).0642

Those are two points from each of the four quadrants.0655

The next example: Graph each point on the same coordinate plane.0659

Let me do these: the first one, point A, is (0,3).0663

Now, be careful--this 0 is my x; that means, on my x-axis, I am going to be at 0, which is right there.0685

And then, for my y (I'll just write out a few of these numbers: 1, 2, 3, 4, 1, 2, 3...OK, let me erase that...-3 and -4; OK)...0698

again, it is 0 for my x, and then 3 on my y; so there is 3 on my y.0725

And that is going to be my point A.0735

For point B, I'll go to -2 on my x and -1 on my y; there is B.0740

C is -5; there is -5 on my x and 0 on my y; that means I am not going to move up or down; I am going to stay there; there is C.0748

And then, D will be 4, and then -6 is all the way down here; so there is point D.0758

And my final example: Point A is (3,1) and B is (0,-5); they both lie on the graph y = 2x - 5.0771

Determine whether each point is collinear with points A and B.0781

OK, if I have my coordinate plane, my x- and my y-axis, my point A is going to be (3,1); there is A.0785

B is going to be (0,-5), right there; there is B.0802

They both lie on the graph y = 2x - 5; so if I draw a line through these points, that is going to be the line for this equation of y = 2x - 5.0812

And you are just going to determine whether each point is collinear with the points A and B.0832

Now, "collinear" means that they are going to be on the same line.0837

So, we are just going to see if these three points (since we know that points A and B are on this line) are going to also be on the line.0841

And if they are, then they will be collinear with the points.0852

For point C, instead of graphing the line and seeing if the point lies on the line, you can just...0858

since you know that this is x and this is your y, you can just plug it into the equation and see if it works.0866

y = 2x - 5: you are just going to plug in -1 for x and 4 for y.0874

So, 4 = 2(-1) - 5: here, this is 4 = -2 - 5; do we know that...since we don't know that these are equal...does 4 equal -7?0880

No, it does not; so this point does not lie on this line; that means that point C is not collinear--this says no.0906

OK, for point D, I am going to also plug in: 7 is my y; 7 = 2(6) - 5.0918

OK, I am going to put a question mark over my equals sign, just because I am not sure if it does yet--I can't see if it equals.0940

This is 12 - 5; 7 = 7, so this is a yes--they are collinear points.0947

And then, my last point, point E: -15 = 2(-5) - 5): put a question mark again.0958

-10 - 5...-15 does equal -15, so this is also a yes; OK.0972

Points D and E are collinear with points A and B, since they are all on the line y = 2x - 5.0984

That is it for this lesson; thank you for watching Educator.com!0998

Welcome back to Educator.com.0000

This lesson is on points, lines, and planes; we are going to go over each of those.0002

First, let's start with points: all geometric figures consist of points.0010

That means that, whether we have a triangle, a square, a rectangle...we have a line...0017

no matter what we have, it is always going to consist of an infinite number of points.0023

A point is usually named by a capital letter, like this: this is point A.0030

(x,y), that point right there, the ordered pair, is labeled A; it is called point A; that is how it is named--point A--by capital letter.0037

Next, lines: a line passes through two points; so whenever you have two points, you can always draw a line through them.0051

So, a line has at least two points; lines consist of an infinite number of points.0061

With this line here, line n, I have two points labeled here, A and B; but a line consists of an infinite number of points.0070

So, every point on this line is one of the infinite numbers; so we have many, many, many points on this line, not just 2.0079

A line is often named by two points on the line, or by a lowercase script letter.0091

The way we label it, or the way we name it: this is an n in script; I can call this line n, or I can call it line AB (any two points).0096

Now, this line has arrows at each end; that means it is going continuously forever, infinitely continuous.0112

It never stops; since it is going in both directions, I can say that this is line AB, or line BA; this can also be BA.0126

And this is actually supposed to go like this, AB; or it could be BA, because it is going both ways.0140

Line AB...now, when we say line AB, then we don't draw a line above it, like this, because,0153

when we say "line," that takes care of it; we don't have to draw the line, because we are saying "line."0160

Line AB or line BA...this can also be line n in script, or AB with a line above it--a symbol.0167

Next, for planes: a plane is a flat surface that extends indefinitely in all directions.0180

Planes are modeled by four-sided figures; even though this plane is drawn like this, a four-sided figure,0187

it is actually going to go on forever in any direction.0196

If I draw a point here, then I can include that in the plane, because the plane is two-dimensional;0202

so the points could be either on the plane...or it might not be.0210

But they are modeled by four-sided figures; and make sure that it is flat.0218

A plane can be named by a capital script letter or by three non-collinear points in the plane.0225

So, this whole plane is called N; we can name this N, by a capital script letter, or by three non-collinear points in the plane.0230

Here are three non-collinear points (non-collinear, meaning that they do not form a straight line):0245

it could be plane N (the whole thing is titled N, so it could be plane N), or it could be plane ABC: plane N or plane ABC.0254

Now, it doesn't have to be ABC; it could be plane BCA; it could be plane CBA, CAB...either one is fine.0266

Now, drawing and labeling points, lines, and planes: the first one here: we have a line.0281

Now, I know that this is kind of hard to see, because there is so much on this slide; but just take a look at this right here.0290

It is just the first part; this line is line n; I don't have two points on this line labeled,0297

so I can't name this line by its points; I can't call it line S; it has to be line n; that would be the only name for it.0308

So, S, a capital letter--that is how points are labeled: S, or point S, is on n, or line n.0318

I could say that line n contains point S, or I can say that line n passes through S, or point S.0329

Even if it doesn't say plane N or point S, just by the way that the letter is written,0342

how you see the letter, you can determine if it is a point, a line, or a plane.0348

The next one: l and p intersect in R; how do we know what these are?0355

It is lowercase and script; that means that they are names of lines, so line l and line p intersect in R.0362

It is just a capital letter, not scripted, so it is just a point, R; so l and p intersect in R; they intersect at point R.0372

l and p both contain R, meaning that point R is part of line l, and R is part of line p.0381

R is the intersection of l and p; line l and line p--R is the intersection of the two lines.0391

The next one: l (here is l, line l) and T...now, this might be a little hard to see,0402

but when this line goes through the plane, this is where it is touching; so think of poking your pencil through your paper.0412

Right where you poke it through, if you leave your pencil through the paper, that point where your pencil is touching the paper--that would be point T.0429

I know it is kind of hard to see, but just think of it that way.0440

So, line l and T, that point, are in plane P--a capital script letter; that is the plane.0443

P contains point T and line l; line l is just going sideways, so if you just drew a line on the paper, then that would be line l.0453

Line m intersects P, the plane, at T; this line right here intersects the plane at that point--that is their intersection point.0465

T is the intersection of m with P; T is a point; point T is the intersection of line m with plane P.0482

Your pencil through your paper--the intersection of a plane with a line--will be a point, and that is point T.0495

The next one: this is a little bit harder to see; I know that it is kind of squished in there.0504

But here we have two planes: this is plane N, and this is plane R.0511

We have a line that is the intersection of R and N; so if you look at this line, this line is passing through plane R,0519

and it is also passing through plane N; and on that line are points A and B.0533

OK, line AB...the reason why this is labeled line AB is because there is no name for this line; so you just have to name it by any two points on the line.0544

So, AB is in plane N, and it is in plane R; this line is in both.0560

N and R, both planes, contain line AB; what does that mean?0575

If this line is part of both planes, that means that the line is the intersection of the two planes.0587

Think of when you have two planes intersecting; they are always going to intersect at a line.0592

It is not going to be a single point, like a line and a plane; two planes intersect at a line.0599

We will actually go over that later; N and R intersect in line AB.0605

The line AB is the intersection of N and R; there are different ways to say it.0614

Let's go over some examples: State whether each is best modeled by a point, a line, or a plane.0621

A knot in a rope: the knot...if I have a rope, and I have a knot, well, this knot is like a point.0627

This one is going to be a point.0637

The second one: a piece of cloth: cloth--a four-sided figure--that would be a plane.0642

Number 3: the corner of a desk: if I have a desk, the corner is going to be a point.0653

And a taut piece of thread; this thread is going to be a line.0664

The next example: List all of the possible names for each figure.0677

Line AB: this line can be line n; that is one name.0680

It can be like that, line AB or line BA; it can also be BA in symbols, like that, or BA this way.0694

The next one: Plane N: this is one way to name it.0716

I can also say plane ABC, plane ACB, plane BAC, plane BCA, CAB, and CBA.0721

There are all of the ways that I can label this plane.0746

Refer to the figure to name each.0757

A line passing through point A: there is point A; a line that is passing through is line l.0760

Two points collinear with point D (collinear, meaning that they line up--they form a line):0776

two points collinear with point D, so two points that are on the same line: points B and E.0785

A plane containing lines l and n: well, there isn't a plane that contains lines l and n,0800

because this line l is not part of plane R.0819

How do we know? because it is passing through, so it is like the pencil that you poke through your paper.0824

It is not on the plane; it is just passing through the plane.0832

So, a plane containing lines l and n is not here.0836

If I asked for two lines that plane R contains, I could say plane R contains lines n and...0842

the other one right here; there is no name for it, so I can say line FC.0866

I can write it like that, or I can say and line FC, like that.0876

OK, the next example: Draw and label a figure for each relationship.0881

The first one is point P on line AB.0890

It is a line...draw a line AB; there is AB, and point P is on the line, so we can draw it like that.0895

CD, the next one: line CD lies in plane R and contains point F.0910

So, I have a plane; line CD lies in plane R (this is plane R) and contains point F; the line contains point F.0918

Points A, B, and C are collinear, but points B, C, and D are non-collinear.0947

OK, that means I can just draw a line first, or I can just draw the points first, points A, B, and C.0953

They are collinear, but points B, C (there are B and C), and D are non-collinear; so I can just draw D somewhere not on the line.0968

OK, so A, B, and C are collinear, but B, C, and D are non-collinear.0979

OK, the next one: planes D and E intersect in n.0985

Now, this is a line, because it is a lowercase script letter.0990

So, here is one plane; here is another plane; let's label this plane D; this could be plane E.0994

And then, where they intersect, right here--that will be line n.1019

OK, that is it for this lesson; thank you for watching Educator.com.1029

Welcome back to Educator.com.0000

This lesson is on measuring segments; let's begin.0002

Segments: if we have a segment AB, here, this looks like a line; we know that this is a line, because there are arrows at the end of it.0009

But if we are just talking about this part from point A to point B, that is a segment,0019

where we are not talking about all of this--just between point A and point B.0027

That would be a segment, and we call that segment AB.0035

And it is written like this: instead of having...if it was just a line, the whole thing, that we were talking about, then we would write it like this.0038

But for the segments, we are just writing a line like that--a segment above it.0048

A segment is like a line with two endpoints.0057

And if we are talking about the measure of AB, the measure of AB is like the distance between A and B.0062

And when you are talking about the measure, you don't write the bar over it--you just leave it as AB.0074

So, when you are just talking about the segment itself, then you would put the bar over it;0080

if not, then you are just leaving it as AB; OK.0084

Ruler Postulate: this has to do with the distance: The points on any line can be paired with real numbers0093

so that, given any two points, P and Q, on the line, P corresponds to 0, and Q corresponds to a positive number,0102

just like when you want to measure something--you use a ruler, and you put the 0 at the first point;0111

and then you see how long whatever you are trying to measure is.0122

In that same way, if I have two points on a number line--let's say I have a point at 2 and another point at 8--0128

then if I were to use a ruler to find the distance between 2 and 8, I would place my 0 here, on the 2.0138

That is what I am saying: the first number corresponds to 0.0146

It is as if this becomes a 0, and then we go 1, 2, 3, 4, 5, 6; so whatever this number becomes, when this is 0--that would be the distance.0149

And when you use the ruler postulate, you can also find the distance of two points on the number line, using absolute value.0166

I can just subtract these two numbers, 2 minus 8; but I am going to use absolute value.0176

So then, 2 - 8 is -6; the absolute value of it is going to make it 6.0186

Remember: absolute value is the distance from 0; so if it is -6, how far away is -6 from 0? 6, right?0192

So, absolute value just makes everything positive.0200

You can also...if I want to find the distance from 8 to 2, it is the same thing: the absolute value of 8 minus 2.0205

OK, if I measure the distance from here to here, it is the same thing as if I find the distance from this to this.0215

That is also 6; so either way, your answer is going to be 6.0223

Another example: Find the distance between this point, -4, and...let's see...5.0230

I am going to find the distance from -4 to +5.0240

I can do the same thing: the absolute value of -4 - 5; this is the absolute value of -9, which is 9.0244

From -4 to 5...they are 9 units apart from each other.0260

And you can also do the other way: the distance from 5 to -4...minus -4...a minus negative becomes a positive, so this is absolute value of 9, which is 9.0266

You don't have to do it twice; I am just trying to show you that you will have the same distance,0284

whether you start from this number and go to the other number, or you start from the other one and you go the other way.0292

That is the Ruler Postulate.0299

The next one, the Segment Addition Postulate: you will use this postulate many, many, many times throughout the course.0304

A postulate, to review, is a math statement that is assumed to be true.0314

Unlike theorems...theorems are also math statements, but theorems have to be proved0321

in order for us to use them, to accept it as true; but postulates we can just assume to be true.0330

So, any time there is a postulate, then we don't have to question its value or its truth.0334

We can just assume that it is true, and then just go ahead and use it.0344

Segment Addition Postulate: if Q is between P and R, then QP + PR = PR.0348

If Q...I think this is written incorrectly...is between P and R...this is supposed to be QR; so let me fix that really quickly.0365

QP + QR = PR: so Q is between P and R--let me just write that out here.0382

If this is P, and this is R, and Q is between P and R; then they are saying that QP or PQ plus QR, this one, is going to equal the whole thing.0390

It is...if I have a part of something, and I have another part of something, it makes up the whole thing.0408

And if PQ + QR equals PR, then Q is between P and R; so you can use it both ways.0414

And it is just saying that this whole thing is...let's say that this is 5, and this is 7; well, then the whole thing together is 12.0425

Or if I give you that this is 10, and then the whole thing is 15, then this is going to be 5, right?0442

That is all that it is saying: the whole thing can be broken up into two parts, or the two parts can be broken up into two things...0456

it just means that, if it is, then Q is between P and R; or if they give you that Q is between P and R--0463

if that is given, that a point is between two other points on the segment, then you can see that these two parts equal the whole thing.0471

That is the Segment Addition Postulate.0480

Find BC if B is between A and C and AB is 2x - 4; BC is 3x - 1; and AC equals 14.0484

Find BC if B is between A and C...let's draw that out: here is A and C, and B is between them.0500

It doesn't have to be in the middle, just anywhere in between those two points.0510

If I have B right here, then we know that AB, this segment, plus this segment, equals the whole segment, AC.0514

So, AB is 2x - 4; and BC is 3x - 1; the whole thing, AC, is 14.0525

I need to be able to find BC; well, I know that, if I add these two segments, then I get the whole segment, right?0541

So, I am going to do 2x - 4, that segment, plus 3x - 1, equals 14.0548

So, here, to solve this, 2x + 3x is 5x; and then, -4 - 1 is -5; that equals 14.0563

If I add 5 to that, 5x = 19, and x = 19/5; OK.0577

And they want you to find BC; now, you found x, but always look to see what they are asking for.0586

BC = 3(19/5) - 1; and then, this is going to be 57/5, minus 1; so I could change this 1 to a 5/5,0602

because if I am going to subtract these two fractions, then I need a common denominator.0625

Minusing 1 is the same thing as minusing 5/5; and that is only so that they will have a common denominator, so that you can subtract them.0631

And then, this will be 52/5; you could just leave it as a fraction.0639

And notice one thing: how these BC's, these segments, don't have the bars over them.0649

And that is because you are dealing with measure: whenever you have a segment equaling its value,0656

equaling some number, some distance, some value, then you are not going to have the bar over it, because you are talking about measure.0661

The next one: Write a mathematical sentence given segments ED and EF.0676

This is using the Segment Addition Postulate; that is the kind of mathematical sentence it wants you to give.0682

ED and EF--that is all you are given, segments ED and EF.0689

Well, here is E; there is an E in both; that means that E has to be in the middle.0696

E has to be here, because since it is in both segments, that is the only way I can have E in both.0707

So then, here, this will be D, and this can be F.0719

Now, this can be F, and this can be D; it doesn't really matter,0722

as long as you have E in the middle, somewhere in between, and then D and F as the endpoints of the whole segment.0726

And to use the Segment Addition Postulate, I can say that DE or ED, plus EF, equals DF; we just write it like that.0737

OK, the Pythagorean Theorem: In a right triangle, the sum of the squares of the measures of the legs equals the square of the measure of the hypotenuse.0758

You probably remember this from algebra: if you have a right triangle...0770

now, you have to keep that in mind; the Pythagorean Theorem can only be used on right triangles;0776

a right triangle, and you use it to find a missing side.0785

You have to be given two out of the three sides--any two of the three sides--to find the missing side.0789

That is what you use the Pythagorean Theorem for--only for right triangles, though.0796

So, a2 + b2 = c2: that is the formula.0799

You have to make the hypotenuse c; this has to be c.0809

Now, just to go over, briefly, the Pythagorean Theorem, we have a right triangle, again.0815

Now, let's say that this is 3; then, if a2 + b2 = c2, then a and b are my two sides, my two legs, a and b.0829

The hypotenuse will always be c; it doesn't have to be c, but from the formula, whatever you make this equal to--0844

the square of the sum of the two sides--has to be...0852

I'm sorry: you have to square each side, and then you take the sum of that; it equals the hypotenuse squared.0858

OK, and let me just go over this part right here.0864

If we have this side as 3, then you square it, and it becomes 9.0870

Now, you can also think of it as having a square right there; so if this is 3, then this has to be 3; this whole thing is 9.0877

If this is 4, if I make a square here, then this has to be 4; this whole thing is 16--the area of the square.0890

And then, it just means that, when you add up the two, it is going to be the area of this square right here.0907

Then, the area of this square is going to be 25, because you add these up, and then that is going to be the same.0917

And then, that just makes this side 5.0925

a2...back to this formula...+ b2 = c2: you just have to square the side,0935

square the other side, add them up, and then you get the hypotenuse squared.0942

Let's do a problem: Find the missing side.0950

I have the legs, the measure of the two legs, and I need to find the hypotenuse.0956

So, that is 4 squared, plus 3 squared, equals the hypotenuse squared; so I can just call that c squared.0961

So, 4 squared is 16, plus 9, equals c squared; 25 = c2.0972

And then, from here, I need to square root both; so this is going to become 5.0983

Now, normally, when you square root something, you are going to have a plus/minus that number;0991

but since we are dealing with distance, the measure of the side, it has to be positive; so this right here is 5.0997

OK, the distance formula: The distance between any two points with coordinates (x1,y1)1010

and (x2,y2), is given by the formula d = the square root1018

of the difference of the x's, squared, plus the difference of the y's, squared.1025

Here, this distance formula is used to find the distance between two points.1035

And we know that a point is (x,y); and the reason why it is labeled like this...1044

you have to be careful; I have seen students use these numbers as exponents.1051

Instead of writing it like that, they would say (x2,y2); that is not true.1056

This is just saying that it is the first x and the first y; so this is from the first point.1061

They are saying, "OK, well, this is (x,y) of the first point; and this is the second point."1065

And that is all that these little numbers are saying; they are saying the first x and first y,1076

from the first point, and the second x and the second y from the second point.1082

x2 just means the second x, the x in the second point.1090

And it doesn't matter which one you make the first point, and which one you make the second point;1094

just whichever point you decide to make first and second, then you just keep that as x2, x1, y2, and y1.1100

Find DE for this point and point D and E; so then, I can make this (x1,y1), my first point;1108

and then this would be (x2,y2)--not (x2,y2); it is (x2,y2), the second point.1119

Then, the distance between these two points...I take my second x (that is 1) minus the other x, so minus -6, squared,1127

plus the second y, 5, minus the other y (minus 2), squared.1143

1 - -6: minus negative is the same thing as plus the whole thing, so this will be 7 squared plus 3 squared.1156

7 squared is 49; plus 3 squared is 9; this is going to be 58.1170

Now, 58--from here, you would have to simplify it.1182

To see if you can simplify it, the easiest way to simplify square roots--you can just do the factor tree.1187

I just want to do this quickly, just to show you.1194

A factor of 58 is going to be 2...2 and 29.1199

Now, 2 is a prime number, so I am going to circle that.1207

And then, 29: do we have any factors of 29? No, we don't.1211

So, this will be the answer; we know that we can't simplify it.1217

The distance between these two points is going to be the square root of 58.1224

Let's do a few examples that have to do with the whole lesson.1234

Find each measure: AC: here is A, and here is C.1238

You can use the Ruler Postulate, and you can make this point correspond to 0.1246

And then, you see what C will become, what number C will correspond to.1251

Or, you can just use the absolute value; so for AC, this right here...AC is the absolute value of -6 minus...C is 2;1259

so that is going to be the absolute value of -8, which becomes 8.1275

BE: absolute value...where is B? -1, minus E (is 9)...so this is the absolute value of -10, which is 10.1284

And then, DC: the absolute value of...D is 5, minus 2.1301

Now, see how I went backwards, because that was DC.1311

It doesn't matter: you can do CD or DC; with segments, you can go either way.1313

So, DC is 5 - 2 or 2 - 5; it is going to be the absolute value of 3, which is 3.1319

The next example: Given that U is between T and V, find the missing measure.1332

Here, let's see: there is T; there is V; and then, U is just anywhere in between.1342

TU, this right here, is 4; TV, the whole thing, is 11.1356

So, if the whole thing is 11, and this is 4, well, I know that this plus this is the whole thing, right?1362

So, you can do this two ways: you can make UV become x; I can make TU;1370

or plus...UV is x...equals the whole thing, which is 11.1377

You can solve it that way, or you can just do the whole thing, minus this segment.1384

If you have the whole thing, and you subtract this, then you will get UV.1391

You can do it that way, too; if you subtract the 4, you get 7, so UV is 7.1395

The next one: UT, this right here, is 3.5; VU, this right here, is 6.2; and they are asking for the whole thing.1407

So, I know that 3.5 + 6.2 is going to give me TV.1417

If I add this up, I get 9.7 = TV.1428

And the last one: VT (is the whole thing) is 5x; UV is 4x - 1; and TU is 2x - 1; so they want to define TU.1440

I know that VU, this one, plus TU--these are the parts, and this is the whole thing.1459

The whole thing, 5x, equals the sum of its parts, 4x - 1 plus (that is the first part; the second part is) 2x - 1.1471

I am just going to add them up; so this will be 6x - 1 - 1...that is -2.1487

And then, if I subtract this over, this is going to be -x = -2, which makes x 2.1499

Now, look what they are asking for, though: you are not done here.1505

They are asking for TU, so then you have to take that x-value that you found and plug it back into this value right here, so you can find TU.1508

TU is going to be 2 times 2 minus 1, which is 4 minus 1, which is 3; so TU is 3.1520

The next example: You are finding the distance between the two points.1537

The first one has A at (6,-1) and B at (-8,0); so again, label this as (x1,y1);1542

this has to be x; this has to be y; you are just labeling as the first x, first y;1554

this is also (x,y), but you are labeling it as x2, the second x, and then the second y.1559

The distance formula is the square root of x2, the second x, minus the other x, squared, plus the difference of the y's, squared.1565

x2 is -8, minus 6, squared, plus...and then y2 is 0, minus -1, squared; this is -14 squared, plus 1 squared.1586

-14 squared is 196, plus 1...and then this is just going to be the square root of 197.1615

And then, the next one: I have two points here: I have A at this point, and I have B at this point.1644

Now, even though it is not like this problem, where they give you the coordinates, they are showing you the coordinates.1653

They graphed it for you; so then, you have to find the coordinates of the points first.1660

This one is at...this is 0; this is 1; this is 2; then this is 2; A is going to be at (2,1), and then B is at (-1,-2)...and -2.1664

So then, to find the distance between those...let's do it right here:1686

it is going to be...you can label this, again: this is (x1,y1), (x2,y2).1691

So, it is -2, the second x, minus the first x, minus 2, squared, plus the second y, -2, minus 1, squared.1704

And then, let me continue it right here, so that I have more room to go across:1722

the square root of...-2 - 2 is -4, squared; and then plus...this is -3, squared;1728

-4 squared is 16, plus...this is 9; and that is the square root of 25, which is a perfect square, so it is going to be 5.1747

The distance between these two points, A and B, is 5.1764

And you could just say 5 units.1768

The last example: we are going to use the Pythagorean Theorem to find the missing links.1774

It has a typo...find...1781

And these are both right triangles; I'll just show that...OK.1784

The first one: the Pythagorean Theorem is a2 + b2 = c2.1789

Here, I am missing this side; I am going to just call that, let's say, b.1801

So, a...it doesn't matter if you label this a or this a; just make sure that a and b are the two sides.1807

a2 is 52; plus b2, equals 132.1816

5 squared is 25, plus b squared, equals 169; and then, subtract the 25; so b2 = 144.1825

Then, b equals 12, because you square root that; so this is 12.1840

The second one: I am going to label this c, because it is the hypotenuse.1851

Then: a2 + b2 = c2, so 62 + 82 = c2.1857

62 is 36, plus 64, equals c2; these together make 100, so c is 10.1868

OK, well, that is it for this lesson; thank you for watching Educator.com.1883

Welcome back to Educator.com.0000

This lesson is on midpoints and segment congruence.0002

We are going to talk about more segments.0007

The definition of midpoint: this is very important, the definition of a midpoint.0010

Midpoint M of PQ is the point right between P and Q, such that PM = MQ; that means PM is equal to MQ.0017

So, if I say that M, the midpoint...it is the point right in the middle of P and Q, so it is the "midpoint," the middle point--this is M...0033

then I can say that PM, this right here, is equal to MQ, because you are just cutting it in half exactly, so it is two equal parts.0049

I can write little marks like that to show that this segment right here, PM, and QM are the same.0058

That is the definition of midpoint; that means that, if PQ, let's say, is 20, and M is the midpoint of PQ, then PM is going to be 10; this is 10, and this is 10.0069

So then, if it is the midpoint, then this part will have half the measure of the whole thing.0084

OK, some formulas: the first one: this is on a number line--that is very important.0093

Depending on where we are trying to find the midpoint, you are going to use different formulas.0101

On the number line, the coordinates of the midpoint of a segment whose endpoints have coordinates a and b is (a + b) divided by 2.0106

So, again, only on a number line: if I have a number line like this, say...this is 0; this is 1; 2, 3, 4, 5;0116

if I want to find the midpoint between 1 and 5--so then, this will be a, and this will be b--0132

then I just add up the two numbers, and I divide it by 2.0140

That is the same...just think of it as average: whenever you try to find the midpoint on a number line, you are finding the average.0145

You add them up, and you divide by 2: so it is just 1 + 5, divided by 2, which is 6/2, and that is 3; so the midpoint is right here.0151

Now, if you are trying to find the midpoint on a coordinate plane, then it is different, because you have points;0171

you don't have just numbers a and b--you have points.0181

So, to find the midpoint with the endpoints (x1,y1) and (x2,y2),0184

you are going to use this formula right here: remember from the last lesson: we also used (x1,y1)0193

and (x2,y2) for the distance formula--remember that, for this one,0200

it is not (x2,y2), because that is very different.0205

This right here, these numbers, 1...it is just saying that this is the first point.0208

And then, (x2,y2): it is the second point, because all points are (x,y).0214

So, they are just saying, "OK, well, then, if this is (x,y) and this is (x,y)..." we are just saying that that is the first (x,y) and these are the second (x,y).0221

You are given two points, and you have to find the midpoint.0231

Then, you just take the average, so it is the same as this formula on the number line--0235

you are just adding up the x's, dividing it by 2, and that becomes your x-coordinate;0242

and then you add up the y's, and you divide by 2.0248

You are just taking the average of the x's and the average of the y's.0251

Remember: to find the average, you have to add them up and then divide by however many of them there are.0254

So, in this case, we have two x's, so you add them up and divide by 2.0260

For the y's, to find the average, you are going to add them up and divide by 2; and that is going to be your midpoint.0266

So, to find the average, you are finding what is exactly in between them.0272

Let's do a few examples: Use a number line to find the midpoint of AB.0279

Here is A at -2, and B is at 8; and this is supposed to have a 7 above it.0287

AB: to find the midpoint, to find the point that is right between A and B, I am going to add them up and divide by 2: so -2 + 8, divided by 2.0298

Again, just think of midpoint as average; so add them up and divide by 2.0317

It is going to be 6; -2 + 8 is 6, over 2; and then 3; so right here--that is the midpoint.0324

You can kind of tell if it is going to be the right answer; that is the midpoint, the point right in the middle of those two.0337

If I got 0 as my answer, well, 0 is too close to A and far away from B, so you know that that is not the right answer.0345

The same thing if you get 5 or 6 or even 7--you know that that is the wrong answer.0353

So, it should be right in the middle of those two points.0358

OK, to find the midpoint of AB here--well, we know that we are not going to use the first one, (a + b)/2,0363

because this is in the coordinate plane, and we have to use the second one, where we have (x1,y1) and (x2,y2).0377

So, in this case, since they don't give us the points, we have to find the points ourselves.0390

So, A is at (1,2); this is 1, 2, 3; and then, B is at (-1,-3).0394

For the midpoint, it is the same concept as finding it on the coordinate plane.0412

When you are finding the average, you are just finding the average of the x's; and then you find the average of the y's.0418

You just have two steps: all I do is take...0424

Now, it doesn't matter which one you label (x1,y1) and which is one is (x2,y2).0429

So then, we could just make this (x1,y1), and this could be (x2,y2); it does not matter.0436

You have to have this thing x and this thing y...both of these thing x's and both of these thing y's.0446

Let's see: x2 + x1, or x1 + x2, is 2, plus -1, divided by 2;0456

that is the average of the x's; comma; the average of the y's is going to be 3 + -3, over 2.0467

Here we have 1/2; and 3 + -3 (that is 3 - 3) is 0; our midpoint is going to be at (1/2,0); so this right here is our midpoint.0479

After you find the midpoint, kind of look at it and see if it looks like the midpoint.0499

A couple more problems: If C is at (2,-1), that is the midpoint of AB, and point A is on the origin, find the coordinates of B.0511

C is the midpoint of AB; so if I have AB there, and C is the midpoint right there (there is C), point A is on the origin; find the coordinates of B.0529

Then C is (2,-1), and then A is (0,0): now this is obviously not what it looks like; it is on the coordinate plane, since we are dealing with points.0543

But this is just so I get an idea of what I am supposed to be doing, because in this type of problem, they are not asking us to find the midpoint.0555

They are giving us the midpoint, and they want us to find B--they want us to find one of the endpoints.0562

So, we know the midpoint; so how do we solve this?0570

The midpoint is (2,-1): well, the formula I know is...how did you get (2,-1)?--how did you get the midpoint?0573

You do x1 + x2; you add up the x's and divide it by 2;0584

and then to find the y-coordinate, you add up the y's, and you divide it by 2.0590

And that is 2, and this is -1; so this is the formula to find the midpoint; this is the midpoint.0599

All of this equals this, and all of this equals that.0610

I can just say, if we make A (0,0)--if this is our (x1,y1)...then what is this going to be? (x2,y2), right?0617

It is as if our coordinates for B are going to be the x2 (this is what we are solving for)...0633

we are solving for x2, and we are solving for y2.0640

Those are the two points that we need.0644

Going back to this--well, we know what x1 was, and we know that this whole thing equals this whole thing.0649

So, I am going to make this a 0, plus x2/2; all of that equals 2.0657

When you found the average of the x's, you got 2; but you just don't know what this value is.0670

Then, if you solve for x2, you get 4; so x2 = 4.0681

Then, you have to do the same thing for the y's; so you add up the y's; this plus this, divided by 2, equals -1.0689

You just write that out; 0 + y2, divided by 2, equals -1.0703

And this is what I am solving for, again; so what is y2? y2 became -2, because you multiply the 2 over: -2.0714

That means that our coordinates for B are (4,-2).0728

So again, you can use the formula to find one of the endpoints, too.0741

If they give you the midpoint, then just use it; you have to use this formula to come up with these values, too.0748

So then, we know that the sum of the x's, divided by 2, is 2; so you do 0 + x2 = 2.0759

And then, the sum of the y's, divided by 2, is -1; so you do 0 + y2, divided by 2, equals -1.0768

OK, we will do another example later.0778

The second one: E is the midpoint of DF; let me draw DF; and E is the midpoint.0782

DE, this, is 4x - 1; and EF is 2x...this must be plus 9...find the value of x and the measure of DF.0799

They want us to find x and the value of the whole thing.0816

Since we know that E is the midpoint, we know that DE and EF are exactly the same.0821

So, I can just take these two and make them equal to each other: 4x - 1 = 2x + 9.0830

And then, you solve for x; subtract it over: 2x equals...you add 1...10...x equals 5.0838

Then that is one of the things they wanted us to find: x = 5.0847

And then, find the measure of DF; how do you find the measure of DF?0853

Well, I have x, so I am able to find DE, or I can find EF.0858

Let's plug it into DE: 4 times 5...you substitute 5 in for x...minus 1; there is 20 - 1, which is 19.0866

If this is 19, then what does this have to be? 19.0879

Just to double-check: 2 times 5 is 10, plus 9 is 19.0883

You can just do 19 + 19, or you can do 19 times 2; you are going to get DF = 38, because it is this times 2...19 + 19...it is the whole thing.0888

Segment bisector: Any segment, line, or plane that intersects a segment at its midpoint.0915

A segment bisector is anything that cuts the segment in half.0922

It bisects the segment, meaning that it cuts it in half; so bisecting just means that it cuts it in half; think of it that way.0931

Here, these little marks mean that this segment and this segment are the same.0939

CD is the segment bisector of AB, because CD is the one that cut AB in half.0948

Now, the one that is doing the cutting--the one that is bisecting, or the one that is cutting in half--is the segment bisector.0959

This bisected the segment AD; this is a line segment, CD; this segment bisector is a segment.0968

It doesn't have to be just a segment; it could be a segment; it could be a line; it could be a plane.0978

In this case, it is a segment; if you draw it out like this, it is still a segment bisector, because it is the segment that was bisected.0985

The bisector can be anything: it can be in the form of a line, too.0999

It can also be in the form of a plane; so if I have (I am a horrible draw-er, but) something like this, and it bisects it right there,1003

then a plane could be a segment bisector, because it intersects the segment at its midpoint, point D.1024

Again, a segment bisector is anything that cuts the segment in half, that bisects it, that intersects it at its midpoint.1035

So, we are going to just talk a little bit about proofs, because the next thing we are going to go over is a theorem.1051

And we have to actually prove theorems.1059

Remember: we talked about postulates--how postulates are statements that we can just assume to be true,1062

meaning that once they give us a postulate, then we can just go ahead and use it from there.1069

Theorems, however, have to be proved or justified using definitions, postulates, and previously-proven theorems.1074

So, whenever there is a theorem--some kind of statement--then it has to be backed up by something.1082

It has to show why that is true; and then you prove it.1090

And then, once it is proven, you can start using that theorem from there on, whenever you need to.1094

Now, a proof is a logical argument in which each statement you make is backed up by a statement that is accepted as true.1102

You can't just give a statement; you have to back it up with a reason: why is that statement true?1109

And that is what a proof is.1115

Now, there are different types of proofs; the one that is used the most is called a two-column proof.1116

But we are going to go over that, actually, in the next chapter.1125

The paragraph proof is one type of proof, in which you write a paragraph to explain why a conjecture for a given situation is true.1128

So, you are just explaining in words why that is true.1140

And a conjecture is an if/then statement: if something is true, then something else.1145

So, actually, you are going to go over that more in the next chapter, too.1153

But that is all you are doing; for a paragraph proof, you are just explaining it in words.1156

So, instead of a two-column proof, where you are listing out each statement, and you are giving the reasons for that,1162

to prove something, in a paragraph proof, you are just writing it as a paragraph to prove that a theorem is true.1168

The first proof that we are going to do is going to be a paragraph proof.1180

And that is to prove the midpoint theorem; so again, this is a theorem; it is not a postulate; so we can't just assume that this statement is true.1186

The midpoint theorem says that, if M is the midpoint of AB (here is AB), then AM is congruent to MB.1196

Now, we know, from the definition of midpoint, that if M is the midpoint of AB, then they have equal measures; that is the definition of midpoint.1208

And that is just talking about the measures of them.1223

But "congruent"--to show that AM is congruent to MB--that is the theorem, and we have to prove that first.1226

Given that M is the midpoint of AB, write a paragraph proof to show that AM is congruent to MB.1241

And then, the only purpose of this proof that we are going to do right now is to prove that this midpoint theorem is true.1250

And then, from there, we can just use it.1260

We are going to always start with the given: given that M is the midpoint of AB--that is the starting point.1265

Let's just write it out: From the definition of midpoint, we know that, since M is the midpoint of AB, AM is equal to MB.1278

Now, that means that AM and MB have the same measure--that AM has the same measure as MB.1306

Then, by the definition of congruence...the definition of congruence is just when you switch it from equal to congruent, or from congruent to equal.1333

So, by the definition of congruence, if AM is equal to MB, then (and I can just switch it over to congruent), AM is congruent to MB.1352

They are congruent segments--something like that.1379

You don't have to write exactly the same thing, but you are just kind of showing that we know1385

that we went over the definition of midpoint, and that is AM = MB.1389

And then, from there, you use the definition of congruence to show that AM is congruent to MB.1393

They are congruent segments--that is what the definition of congruence says.1401

Now that we have proven that the midpoint theorem is correct, or is true, then now, from now on,1410

for the remainder of the course, you can just use it whenever you need to--the midpoint theorem.1419

Extra Example 1: Use a number line to find the midpoint of each: BD.1427

We are going to find the midpoint of BD; again, to find the midpoint, you want to find the point that is right in the middle.1434

So, you are going to add up the two points and divide it by 2; so it is -2 + 9, divided by 2; that is 7/2.1442

You can leave it like that, or you can just write 3 and 1/2.1456

Between -2 and 9, it is going to be right there: three and a half.1465

CB: you know, I shouldn't write it like this, because it looks like BD...this looks like distance.1475

You can just write the midpoint of BD: so just be careful--don't write BD; don't write it like that.1496

Just write midpoint of BD, or...I am just going to write the number 2, just so that we know that that is the midpoint of CB.1505

CB is right here; now, you can go from B to C, or we can go from C to B.1514

It doesn't matter; if I go from C to B, 4 + -2, over 2, is going to be 2 divided by 2, which is 1.1520

That means the midpoint from here, C, to B, is 1, right there.1535

You can also check; you can count; this is three units, and then this is three units, so it has to be the same.1545

And then, number 3: AD: A is -5; D is 9; divided by 2...this is 4, divided by 2, which is 2.1553

Here is -5, and here is 9; the midpoint is right here, between those two...from here to here, and from here to here.1568

OK, draw a diagram to show each: AC bisects BD.1586

That means that this is the one that is doing the bisecting; this is the segment bisector.1591

This is the segment that is getting bisected: so BD...here is BD.1597

Now, it doesn't have to look like mine, just as long as BD is getting bisected, or AC is intersecting BD at BD's midpoint.1606

You can just...if I say that this is the midpoint, then AC is the one that is cutting it like this; it is going to be A, and then C (it cuts it)...1619

OK, the next one: HI is congruent to IJ.1638

HI: see how this is congruent...this is part of the midpoint theorem...HI and IJ.1644

Here is HI; I has to be that in the middle; and J; so HI is congruent to IJ, and that is how you would want to show it; OK.1657

The next one: RT equals half of PT--let me just draw another segment; RT is equal to half of PT.1671

Now, look at what they have in common; here is T, and here is T.1683

Now, if RT is equal to half of PT, that means that you have to divide PT by 2 to get RT.1689

That means that the whole thing is going to be PT, because, if you have to divide it by 2, you cut it in half, meaning at its midpoint.1701

And that is going to be RT; that means R is right here--the midpoint.1712

Again, here is PT; so for example, if PT is 12 (say this whole thing, PT, is 12),1720

then you divide it by 2, or you multiply it by 1/2; then you get 6; that means RT has to equal 6.1728

And you don't have to write the numbers; just draw the diagram.1742

You could have just left it at P-R-T, or just showed it, maybe, like this.1747

The next example: Q is the midpoint of RS; if two points are given, find the coordinates of the third point.1754

RS's midpoint is Q; I'll show that; if two points are given, find the coordinates of the third point.1768

R is here; R is at (4,3); S is at (2,1); you have to find the midpoint.1785

To find the midpoint, you are going to take the sum of the x's, divided by 2, and that is going to be your x-coordinate.1799

So, x1 + x2, divided by 2, equals your x-coordinate for the midpoint.1808

4 + 2, divided by 2...and then you are going to take the y's, 3 + 1, divided by 2;1817

that is going to be 6 divided by 2, which is 3; so Q is 3, comma...4 divided by 2 is 2;1831

it is going to be at (3,2); there is the midpoint for that one.1844

The next one: let me just redraw R-Q-S...Q...on this one, they give you the midpoint, and they give you this.1851

This right here is going to be...you could make this (x1,y1), or (x2,y2).1872

And then, when you write it out, it is going to be -5 + x2 over 2, and then 2 + y2, divided by 2.1882

Now, we know that all of this equals...the midpoint is (-2,-1).1904

That means that these equal each other and these equal each other.1912

This is going to give you -2; and this is what we are solving for.1915

I can just make this thing equal to this thing: -5 + x2, over 2.1927

Now, this is not x2; be careful of that.1933

That equals -2; this is going to be -5 + x2 equals...you multiply the 2 over to the other side...-4.1936

You add the 5 over; so x2 = 1.1948

I am going to do the y over here; then you make this equal to -1; it is 2 + y2, over 2, equals -1.1956

You multiply the 2 over; 2 + y2 = -2; y2 = -4.1967

All right, S is going to be (1,-4); that is to find this right here.1977

With this one, we had to find Q, the midpoint; and with this one, we had to find S.1995

The last example: EC bisects AD; that means that EC is the one is doing the bisecting, and AD is the one that got bisected.2002

That means that AD is the one that got cut in half at C.2018

That means that this whole thing, AC, and CD are congruent.2022

EF bisects...this is supposed to be a line...and let's make this F, right here; that means that line EF bisects AC at B.2030

That means that AC is bisected; B is the midpoint.2050

For each, find the value of x and the measure of the segment.2056

That means AC and CD are the same, and then AB and BC are the same.2060

AB equals 3x + 6; BC equals 2x + 14; they want you to find AC.2074

Again, AB and BC have the same measure; so I can just make them equal to each other...it equals 2x + 14.2087

Subtract the 2x over here; subtract the 6 over there; you get 8.2101

And then, find the value of x and the measure of the segment.2109

The segment right here, AC, is AB + BC; or it is just AB times 2, because it is doubled.2113

So, AC...here is my x, and AC equals...we will have to find AB first--or we can find BC; it doesn't matter.2126

AB is 3 times 8, plus 6; that is 24; that is 30.2142

24 plus 6 is 30; and then, AC is double that, so AC is 60.2152

AB is 30; that means that AC is 60.2161

AD, the whole thing, is 6x - 4; AC is 4x - 3; and you have to find CD.2172

Now, remember how EC bisected AD; that means that AD is cut in half; C is the midpoint.2184

AD, the whole thing, is 6x - 4; and then, AC is 4x - 3.2192

I can do this two ways: AC is 4x - 3--that means DC, or CD, is 4x - 3.2198

So, I can just do 4x - 3 + 4x - 3, or I can do (4x - 3) times 2; or we can do the whole thing, AD, 6x - 4, minus this one.2208

I am just going to do 2 times (4x - 3), because AC is 4x - 3, and AD is double that.2224

Now, even though this is 60, so you might assume that this whole thing is 120; this is a different problem.2238

So then, it is not going to have the same measure.2244

2 times AC equals AD, which is 6x - 4.2251

If we continue it here, it is going to be 8x (and this is the distributive property), minus 6, equals 6x minus 4.2261

This becomes 2x; if you add the 6 over there, it equals 2; x = 1.2275

So then, here is the x-value; and then, they want you to find CD, this right here.2283

Since CD is the same thing as AC, I can just find AC.2291

AC is 4 times 1, minus 3; so AC is 4 minus 3, is 1; so CD is 1; AC is 1; CD is 1.2298

OK, the last one: AD, the whole thing, is 5x + 2; and BC is 7 - 2x; find CD.2325

Now, this one is a little bit harder, because they give you AD, the whole thing, and they give you BC.2341

Now, remember: if C is the midpoint of AD, that means that this whole thing and this whole thing are the same.2353

Let's look at this number right here; if this is 60, then this will also be 60.2367

That means that the whole thing together is going to be 120.2374

Then this right here is 30; this right here is 30.2380

It is as if you take a piece of paper and you fold it in half; if you fold it in half, that is like getting your C, your midpoint.2389

Then, you take that paper, and you fold it in half again; so then, that is when you get point B.2399

So, now your paper is folded into how many parts? 1, 2, 3, 4.2405

One of these, AB...if you compare AB or, let's say, BC, to the whole thing, AD, then this is a fourth of the whole thing.2418

So, this into four parts becomes the whole thing; so you can take BC and multiply it by 4, because this is one part, 2, 3, 4.2429

BC times 4: 7 - 2x...multiply it by 4, and you get the whole thing, AD.2446

This is 28 - 8x = 5x + 2; I am going to subtract the 2 over; then I get 26; I am going to add the 8x over, and I get 13x; so x is 2.2456

And then, find CD: CD was 2 times BC, because BC with another BC is going to equal CD.2478

I am just going to find BC: 7 - 2(2)...just plug in x...so 7 - 4 is 3.2492

If BC is 3...now, these numbers right here; that was actually for number 1;2504

those are the values for number 1, and then I just used it as an example for number 3.2511

But don't think that these are the actual values (30 for BC, and then 120 for the whole thing); this is a different problem.2516

BC was 3; then what is CD? CD is 2 times BC--it is double BC, so CD is 6.2526

OK, well, that is it for this lesson; thank you for watching Educator.com.2542

Welcome back to Educator.com.0000

This next lesson is on angles.0002

OK, first, let's go over what an angle is: an angle is a figure formed by two non-collinear rays with a common endpoint.0007

Let's go over what a ray is: a ray is a segment like this, with one endpoint and one end extending indefinitely.0021

It is like part of it is a line, and part of it is a segment--one endpoint and one end going continuously.0035

An angle is a figure when we have two rays together, like this here, with a common endpoint.0048

Here is one ray, and here is another ray.0059

Now, just to go over rays a little bit: if you have a ray like this, then to write it...0063

now, we know that, if we have a line, we write it like this, using symbols...0072

if we have a ray, then you are going to write it like this, with one arrow.0078

Now, be careful: you cannot write it like this--you can't do that, because the endpoint is at A.0082

So then, the arrow has to be pointing to the point that is closer to the arrow.0095

So, if it is AB, it is going this way; so then you have to point the arrow that way: AB.0101

It can't be BA, because that is showing you that that is going the other way.0106

Now, also, don't write it like that, either--the direction has to be going to the right.0110

This is the right way; not like that, and not like that.0120

Opposite rays are when you have two rays that extend in opposite directions.0130

If they go in opposite directions...they have a common endpoint; one ray is going this way; the other ray is going this way.0137

Those are opposite rays: you have two rays: one is going to the right, and one is going to the left.0147

They form a line, because they are going in exactly opposite directions.0153

So then, opposite rays are two rays that extend in opposite directions to form a line.0159

Here is a diagram of an angle; now, this angle right here, where the two rays meet, the common endpoint is called the vertex.0166

And each of these rays in the angle are called sides.0187

This is a side, and then this is a side; this is a vertex, and these are sides, of an angle.0193

List all of the possible names for the angles: here, this is actually made up of three angles.0204

We have this angle right here, and I can just label that angle 1; this is another angle right here, and then this big angle.0214

To list all of the names, you are going to have to look at the different angles.0229

Depending on what angle you are looking at, you are going to call it by a different name.0236

This first one right here is going to be angle ABD; make sure that the middle letter is the vertex.0240

This has to be angle ABD; if you do angle ADB, that is going to look like that; ADB is like this, and that is not the angle, so it has to be angle ABD.0253

You can also say angle DBA, as long as the vertex is in the middle.0267

This can also be angle 1, because this whole angle is labeled as angle 1.0277

That is the first one; and then the next one can be angle DBC (again, with the vertex in the middle), angle CBD, or angle 2.0290

And then, the big one...now, if I just had an angle like this, just a single angle, and the vertex...0307

let's say this was different...EFG: if I had an angle like that, this can be angle angle EFG, angle GFE, or angle F.0314

It could be angle F, even though F is just the vertex; as long as you have only one angle--0333

this vertex is only for a single angle--then you can name the angle by its vertex, so this can be angle F.0341

In this case, I have three different angles here, so I can't label this whole...0351

even if I am talking about the big one, the whole thing, I can't label it as angle B, because angle B...this is a vertex for three different angles.0358

So, I cannot label it angle B; so instead, you have to just list it all out.0369

You are going to just say angle...the big one is label ABC; angle CBA; and that is it--those are the only two names for that big one.0375

It is not by 1 and 2; the 1 is for this angle, and then the 2 is for this angle.0390

An angle separates a plane into three parts; if I have an angle (actually, let me just draw it out--0402

there is my angle), then it separates it into its interior (the interior is the inside--that is all of this right here, on the inside),0414

and then the exterior (which is all of this, the outside of the angle), and then on the angle--the angle itself.0428

So, when you have an angle, there are three different parts: the inside, on the angle, and outside the angle.0441

And just so you are more familiar with these words: interior/exterior is inside/outside.0451

Angles are measured in units called degrees: so if you have this angle right here, this angle could be 110 degrees.0458

The number of degrees in an angle is the measure.0475

If I want to say that this angle is angle ABC, then I can say the measure (and m is for the measure) of angle ABC is 110.0480

That is how you would write it: the measure...and you write the m in front of the angle, angle ABC (what you name it) only when you are giving the measure.0499

The Protractor Postulate: from the last lesson (the last lesson was on segments), remember: we went over the Ruler Postulate.0519

This one is the Protractor Postulate; it is the same concept, but then, because it is an angle, you are not using a ruler; you are using a protractor.0529

And again, a postulate is any statement that is assumed to be true.0537

This is the protractor postulate--this whole thing right here, the statement.0542

Once we go over it, we can assume that it is true, because it is a postulate.0546

Given AB and a number r between 0 and 180, there is exactly one ray with endpoint A extending on either side of ray AB0552

such that the measure of the angle formed is r; all that that is saying is that,0567

if you have this ray right here (this is A; this is B), now, if you have a ray AB, there is only one ray0575

or a single ray that you can draw to get a certain angle measure.0597

If I want an angle measure of 80 from AB, then there is only one ray that I can draw that is going to give you an angle measure of 80.0607

You can't draw two different angles; but it is going to be on both sides--so it can be this side,0619

or I can have an angle on...this is AB...then it can go on the other side, too; so this also can be AB.0624

That is all that they are saying: just for this, if you draw a ray like this to make it 80, you can't draw a different type of ray to also make it 80.0634

It is going to be something else.0645

So, for a number r between 0 and 180, there is only one ray that you can extend from this ray right here to give you that angle measure.0647

And the Protractor Postulate is as if you put this at 0 on your protractor, if you have your protractor like that;0661

and then here is your protractor, and then you have all of your angle measures; that is how you would read it.0677

And then, whatever this says right here--this number--that is going to be your angle measure of this.0688

Make sure that this endpoint is at the 0 of the protractor.0694

Angle Addition Postulate: from the last lesson, the segments lesson, we went over the Segment Addition Postulate.0702

The Segment Addition Postulate was when we had a segment that was broken down into its parts.0711

For the Segment Addition Postulate, remember, you had...this is AB...B is just anywhere in between A and C;0720

then we can say that AB + BC equals AC; so this plus this equals AC.0730

In the same way, we have the Angle Addition Postulate.0741

And all that it is saying is that, if R is in the interior (remember, interior is inside) of angle PQS, the measure of angle PQR0748

plus the measure of angle RQS equals the measure of angle PQS.0764

Measure means that you are talking about degrees, the number of degrees in the angle.0776

If the measure of angle PQR, let's say, is 30 (let's say this angle measure is 30),0785

and the measure of angle RQS is, say, 50; then the measure of angle PQS:0794

you just add them up, and then it is going to be...the whole thing is 80 degrees.0805

That is just the Angle Addition Postulate; in the same way, remember, if this was 3, and this was 5, then AC...you add them up, and you get 8.0812

It is the same thing--the Angle Addition Postulate.0824

You can also say the other way around: if the measure of this angle, plus the measure of this angle,0828

equals the measure of the whole thing, then R is in the interior of angle PQS.0837

Classifying angles: let's go over the different types of angles.0852

An acute angle is any angle that is less than 90 degrees; less than 90 looks like that--an acute angle.0857

A right angle is any angle that measures perfectly 90 degrees, like that.0871

And then, an obtuse angle is any angle that is greater than 90, like that.0882

This is a small angle, a right angle, and a big angle--an obtuse angle.0893

Angle bisector: we also went over a segment bisector.0905

A segment bisector was when you have a segment (or...it could be a segment; it could be a line;0909

it could be a ray; it could be a plane)...anything that cuts the segment in half; it intersects the segment.0919

Let's draw a line: this line intersects this segment at its midpoint.0929

That means that this line is called the segment bisector, because it is bisecting this segment; it is cutting it in half.0936

The same thing happens with an angle bisector: it could be a segment, ray, or line that divides the angle ABC into two congruent angles.0944

This ray, ray BD, is an angle bisector; BD, that ray, is the angle bisector of angle ABC.0955

Again, the bisector is whatever is doing the cutting, the dividing; it has to divide it into two congruent parts; and the same here.0971

This is a segment bisector; this would be the angle bisector, because it is intersecting the angle at exactly its middle.0985

Then, if this is 30, then this has to be 30.0996

Angle relationships: adjacent angles are angles that are next to each other with a common vertex and a side.1006

Angles 1 and 2 are adjacent angles, because they share a side, which is this right here, and a vertex.1028

If I have an angle like this, and then an angle like that, 1 and 2, even though they share a common side,1037

these angles would not be adjacent, because they don't share a vertex.1049

It has to be a side and a vertex; so this is not adjacent angles; these would be adjacent angles.1056

Again, two angles that share a side and a vertex are adjacent angles.1063

Vertical angles: if you have two lines that are intersecting each other1070

("intersecting," meaning that they cross each other--they meet, right here), then they form vertical angles.1077

Vertical angles, when they cross, would be the opposite angle; so this angle, right here, and this angle are vertical angles.1087

Non-adjacent means that they are not next to each other; they are not sharing a side and a vertex.1096

Even though vertical angles share a vertex, they are not sharing a side.1102

The two sides of this angle are this and this; the two sides of this angle are this and this.1106

They are not adjacent; they are non-adjacent angles formed by intersecting lines.1117

In this one right here, 1 and 2 would be vertical angles.1124

Now, this and this are also vertical angles; so then, in this one, 3 and 4 are also vertical angles.1127

There are two pairs of vertical angles when two lines are intersected; always, there are always vertical angles involved.1140

If I have an angle that looks like that, these are not considered...1150

because this is not a line, these are not vertical angles, and these are not vertical angles.1161

They have to be straight lines that are intersecting.1175

Linear pair: a linear pair are adjacent angles that form a line by opposite rays.1181

Remember: opposite rays form a line, so adjacent angles are two angles, like this, angles 1 and 2.1191

Here is angle 1, and here is angle 2; see how they form a line?1205

They are also adjacent, because they are sharing a side and a vertex.1213

A linear pair is two angles that just form a line; a linear pair is the pair of angles that forms a line.1218

Right angles formed by perpendicular lines have perpendicular lines; they are perpendicular;1233

then we know that this is 90 degrees; then that means that this is 90 degrees,1245

because we know that a straight line measures 180 degrees.1253

This is also 90, and this is 90; we are going to go over this again later, but perpendicular lines form right angles.1262

Supplementary angles are two angles that add up to 180, two angles whose measures have a sum of 180.1276

Supplementary is 180; complementary angles are two angles that add up to 90.1290

It has to be two angles that add up to 90 and two angles that add up to 180; this is 90.1302

Now, if I have an angle like this (let's say that this is 120 degrees), then I can say...1310

OK, if I ask you for the angle that makes it add up to 180, that is called the supplement.1325

This is 60 degrees, because these two angles together...if they are supplementary angles, then they have to add up to 180.1338

Two angles (1,2) that add up to 180...1347

If I want to compare them to each other, then I can say, "This is the supplement of this, and this is the supplement of that."1357

So, if I ask you, "What is the supplement of 120?" the answer will be 60.1364

"What is the supplement of 60?" 120; but together, they are supplementary angles.1373

The same thing for complementary: I have two angles; let's say this is 50, and this is 40; they are complementary angles; together they add up to 90.1380

But I can say that this is the complement of 40, and 40 is the complement of 50; together, they are complementary angles.1400

So again, supplementary is 180, and complementary is 90.1411

Sometimes, just be careful not to get confused between complementary and supplementary--which one is 90 and which one is 180.1418

Complementary starts with a C; it comes before S; C is before S, and then 90 is before 180; that is one way you can remember it.1428

Complementary is 90; supplementary is 180; C before S, 90 before 180.1440

Let's do a few examples: the first one: State all possible names for angle 1.1450

Here is angle 1; all of the possible names are...(now, I am not going to write measure of angle,1457

because I am not dealing with its degree measure)...just naming them, you are going to write1467

angle AFB, angle BFA (make sure that the vertex is in the middle); it is also angle 1.1474

Is that all? That is all; I cannot say angle F, because angle F...1493

that is the vertex of so many different angles that you can't use it to name any of those angles.1499

So then, that would be all--just those three.1508

Number 2: Name a pair of adjacent angles and vertical angles.1512

Adjacent angles would be...we can say that there are a lot; as long as they share a vertex and a side...1518

I can say that angles 1 and 2 are adjacent angles; angles 2 and 3; angles 3 and 4;1530

I can also say angle 4 and angle EFA, since this one doesn't have a number; I am going to say angle...let's see, 2...and angle 3.1540

Now, remember: angle 2 and angle 4 are not adjacent, because, even though they share a vertex, they don't share a side.1556

And for vertical angles, I can say angle AFE and angle...what is vertical to AFE?1566

It has to be angle BFC, or angle 2.1591

Be careful: right here, angle 4 and angle 1 are not vertical, because it has to be straight intersecting lines that form the two vertical angles.1599

So, you can say that 4 and 3 together are vertical to angle 1, because it is formed by the same sides, the lines; but not 4 and 1, and not 4 and 2.1613

The third one: Make a statement for angle EFC, using the Angle Addition Postulate.1628

The Angle Addition Postulate: remember, that means that this angle plus the measure of this angle is going to equal the measure of the full angle.1640

An the Angle Addition Postulate is different than an angle bisector; you can still apply the Angle Addition Postulate,1653

but for an angle bisector, it has to be cut in half; the angle has to be cut into two equal parts.1665

The Angle Addition Postulate could be like this, and you can say this, plus this small part, equals the whole thing, the whole angle.1672

That is the Angle Addition Postulate; I am going to say that the measure of angle EFD,1686

plus the measure of angle DFC, equals the measure of angle EFC; and that is my statement.1699

Number 4: Name a point in the exterior of angle AFE.1717

Angle AFE is right there; a point on the exterior...this is the interior, so it is anything outside of that.1724

I can say point B; I can say point C; or I can say point D.1734

I am just going to write point C; that is on the exterior.1737

The next example: Name a pair of opposite rays.1747

Opposite rays would be two rays with a common endpoint going in opposite directions.1754

I can say rays CF and CA...now, be careful; I can't say AC.1761

I can't say AC, because the ray is going this way, so I have to label it as CA, going towards that.1777

If the measure of angle ACB is 50, and measure of angle ACE, the whole thing, is 110, find the measure of angle BCE.1787

This is what I am looking for, x; that is the angle measure that I am looking for.1808

This is using the Angle Addition Postulate; so this, the measure of this angle, plus the measure of this angle, equals the whole thing.1813

So, the measure of angle ACB, plus the measure of angle ECB, equals the measure of angle ACE.1822

This will be 50; 50 plus the measure of angle ECB equals 110.1843

Subtract the 50; so the measure of angle ECB (or BCE--the same thing) equals 60 degrees.1855

Number 3: Name two angles that form a linear pair.1870

This one was to name a pair of opposite rays, two rays that form a line; this is two angles that form a linear pair.1878

That is a pair of angles that form a line; so I can say this angle right here and this angle right here,1889

angle FCD and angle DCA: angle ACD and angle DCF, or FCD.1899

Draw angles that satisfy the following conditions: Number 1: Two angles that intersect in one point.1928

We just need two angles that intersect in exactly one point.1935

I can draw an angle like this, and I can draw an angle like this; it doesn't matter, as long as they intersect at one point.1942

If it asks for two angles that intersect in four points, then that would be like this...1952

or this actually is two points; and then, if you went the other way, then you can just do four: so 1, 2...I meant 2.1963

The next one, measure of angle DEB plus the measure of angle BEF, equals the measure of angle DEF.1974

This is the Angle Addition Postulate; I know that this is going to be the whole thing.1987

And then, E has to be the vertex; then D and F...the measure of angle DEB...DE, and then, that means that B is going to be in the interior of the angle.1996

And then, that, plus the measure of angle BEF, equals the measure of angle DEF.2011

Your diagrams might be a little bit different than mine; but as long as you know that this plus this...2015

as long as B is in the middle of this angle, and E is the vertex, then that is fine.2022

Ray AB and ray BC as opposite rays: opposite rays means two rays that go in opposite directions to form a line.2029

There is A, B...I know that this has to be A, because that is how it is written.2041

A has to be at the endpoint; that and BC as opposite rays...well, I can do B, and then C like that, somewhere here.2049

Or, if you want to just draw it longer...it can be like that, or this will probably be AC; and then in that case, it would just be like...2066

A is a common endpoint; AB is going that way, and then AC is going this way.2089

And then, the last one: two angles that are complementary:2096

you can draw any two angles, as long they add up to 90 degrees (complementary, remember, is 90).2099

I can draw it like that; this could be 45 and 45; they are complementary angles.2110

I can draw two angles separately, maybe like that, and then, say, if this is 60, then I have to draw a 30-degree angle.2123

As long as they add up to 90...it is any two angles that add up to 90.2138

OK, the fourth example: BA and BE are opposite rays, and BC bisects the measure of angle ABD.2145

That means that this is a line, because they are opposite rays; so they are saying that it forms a line.2165

BC bisects this angle ABD; that means, since it bisects it, they are the same measure--they are equal.2171

If the measure of angle ABC equals 4x + 1, and the measure of angle CBD equals 6x - 15, then find the measure of angle CBD.2183

We have these two angles; now, when you draw a little line like that, that just means that they congruent to each other--they are equal.2197

And then, if you have two other angles that are not the same as that, then...2211

let's say these two angles are the same; then you can just draw these two, because you did one for each of these,2219

to show that all the angles that you drew one for are congruent.2226

Then, the next pair of congruent angles--you can just draw two.2230

The measure of angle ABC plus the measure of angle CBD is going to be the measure of angle ABD.2239

But then, they are actually wanting you to find the measure of angle CBD; and this is the angle bisector.2247

I can just make them equal to each other; let me just solve it down here.2254

4x + 1 is equal to 6x - 15; I need to solve for x.2258

So, if I subtract the 6x over, I get (let's write out the answer)...-2x; subtract the 1; and I get -16, so x = 8.2267

But they want you to find the measure of angle CBD; that means that, once you find x, you have to plug it back into to the measure of angle CBD.2286

That is 6(8) - 15; this is 48 - 15; that is 33; so the measure of angle CBD equals 33.2299

Number 2: the measure of angle DBC is 12n - 8; the measure of angle ABD, the whole thing, is 22n - 11; find the measure of angle ABC.2324

They give you this whole thing right here, and they give you DBC, and they give you ABD; and they want you to find this angle right here.2346

Since we know that this and this are the same--they have the same measure--this whole thing would be two times one of these.2359

So, if this is 10, then this has to be 10; then the whole thing is 20.2370

I can do 12n - 8, plus 12n - 8 (because this is also 12n - 8--the same thing), equals 22n - 11, the whole thing.2378

Or you can just do 2 times 12n - 8, because it is just this angle, times 2, equals ABD.2390

So, I am just going to do that; number 2 is 2(12n - 8) = 22n - 11.2400

This is going to be...I will use the distributive property...this is 24n - 16 = 22n - 11.2414

If I subtract this, I get 2n; add it over; I get 5; so n = 5/2.2425

And then, n is 5/2, and then we have to find the measure of angle ABC.2438

Now, they don't give me something for ABC; but as long as I find what DBC is, then that is the same measure.2450

I just do 12...substitute in that n...minus 8; this becomes 6; 6 times 5 is 30, so 30 - 8 is 22.2461

That means that the measure of angle ABC is 22.2480

The last one: If the measure of angle EBD, this one right here, is 115, find the measure of the angle supplementary to angle EBD.2493

Supplementary: it is asking you to find the supplement of this angle.2508

Remember: supplementary is 180, so find two angles that add up to 180; it is 115 + something (which is x) is going to add up to 180.2513

You are going to subtract it: x is equal to 65 degrees.2530

This will be the supplement of angle EBD.2539

All right, that is it for this lesson; we will see you next time.2549

Thank you for watching Educator.com!2552

Welcome back to Educator.com.0000

For the next lesson, we are going to go over polygons.0002

We are going to talk about the different types of polygons and the interior and exterior angles of polygons.0004

First, let's talk about what is a polygon and what is not a polygon.0012

A polygon is a closed figure, formed by coplanar segments, such that the sides are non-collinear,0018

and each side intersects exactly two other sides at their endpoints.0027

Basically, a polygon ("poly" meaning many) is a closed shape (meaning it has to close), each side being a straight lines, and where no sides overlap.0033

As long as we have a shape that is closed (nothing open--nothing can get through), with no overlapping, and...0055

like this one...see how it is not straight...these are examples of polygons, and these are not.0069

Now, it is OK if polygons look funny; if they look like this, that is OK.0078

As long as it is a closed figure, each side is a straight line segment, and none of the sides overlap, then it is a polygon.0082

So, here, because of that it is not a polygon; because of the overlap, it is not a polygon;0094

and because this side right here is not straight, that is not a polygon.0101

The two types of polygons are convex and concave; a convex polygon is when all of the sides are on the outside of the shape.0109

What that means...maybe if I explain "concave," it will be easier to understand.0125

A concave polygon is when two sides go in towards the center of the polygon.0132

See how, right here, these two sides are angled towards the center; that would make this concave.0140

The same thing happens here: we have this angle going towards the center.0148

Think of a cave, like a mountain, or in the rocks; see how it goes inwards, and it creates a little cave?0153

So, any time it does that, it is a concave polygon.0163

If it doesn't, then it is a convex; so all of the angles are pointing away from the center.0166

All of these angles are pointing away from the center, and that is convex.0175

And this explanation here: No line containing a side of the polygon contains a point in the interior of the polygon.0184

It just means that, if you were to draw each of these sides or extend them into lines, it is not going to cut through the polygon.0193

If we make this into a line, it is not going to cut through the inside of the polygon.0211

The same thing happens here, and the same thing here.0216

With this one, however, if I draw a line, see how it cuts in the polygon; that is what it means--that is what this explanation is saying.0221

The easiest way to remember: just think of the cave--it is creating a little space right there, like a cave, so they are concave polygons.0235

Now, a regular polygon is a polygon with all sides congruent and all angles congruent; it is equilateral, and it is equiangular.0246

That shows that it is equilateral, and this shows that it is equiangular.0262

Any time that it is equilateral and equiangular, it is a regular polygon.0269

And in order for it to be equilateral and equiangular, it has to be convex; you can't have a concave polygon that is equilateral and equiangular.0281

Maybe it could be equilateral, but not equiangular.0290

The interior angle sum theorem is to figure out the sum of all of the angles inside the polygon.0294

If I have a triangle (which is actually going to be the first polygon that we are going to use), I have three angles in the triangle.0304

The interior angle sum theorem is going to give me the total, or the sum, or the angle measure, of all three angles combined.0318

So, if I have a quadrilateral (four angles), what do all four angles of the polygon add up to?0329

First, let's start with triangles; a triangle has three sides...number of triangles: a triangle only has one triangle.0337

We are going to talk about this in a little bit, but the number of triangles would just be one.0351

The sum of angle measures: we know that all three angles of a triangle add up to 180.0356

Next will be a quadrilateral, a four-sided polygon: number of sides: 4.0368

Now, if I have a quadrilateral, I have two triangles; so the sum of the angle measures is going to be 360.0377

I know that all of the angles added up together in a quadrilateral are going to add up to 360.0394

And then, a pentagon is the next one: it has 5 sides; the number of triangles is going to be 3 (let's see if I can draw this: that would be 1, 2, 3).0400

The sum of the angle measures: here, every time we add a triangle...every time we have one more side,0427

it is like we are adding another triangle in the polygon, and then we add another 180,0436

because, for every triangle that exists in the polygon, there is an additional 180.0442

We always start with the triangle, because that is the first polygon.0451

Then, when you get to quadrilateral, the next one, it is going to be plus 180.0454

Then, to get to the next one, we are going to do + 180, which is going to be 540;0462

so the angle sum of a pentagon is going to be 540 degrees.0471

How about the next one, which is a hexagon?--6 sides: 1, 2, 3, 4...0480

Again, we just add 180; it is going to be 720; and so on.0498

For each triangle that exists, again, it is going to be 180 degrees.0511

But what if I ask you for a 20-sided polygon--what is the interior angle sum of a 20-sided polygon?0516

There is a formula to go with this, and that is right here.0527

Because a triangle has 3 sides, but only one triangle exists; that is 180.0533

For every additional triangle, it is going to be an additional 180; so here, isn't this 2 times 180?--because it is 180 + 180, which is 360.0542

Here, from a 5-sided polygon, there are three triangles that exist, so isn't that 3 times 180 (180 + 180 + 180, which is 540)?0555

So, it is 180 times the 3; here, there are four triangles, so I have to do 4 times 180.0568

So, if I want to find a 20-sided polygon, how many triangles exist?0577

Well, look at the pattern: 3:1, 4:2, 5:3, 6:4, 20...it is 2 less, so it is going to be 18.0588

Now, again, this is to get 360 here; so we just do 2 times 180, which is going to equal 360; 3 times 180...0601

the number of triangles times 180...4 times 180 was 720.0615

So here, all I have to do is multiply 18 times 180.0621

So first, I have to figure out how many sides I have; this is going to be n.0629

And then, subtract the 2 to figure out how many triangles exist in that polygon; and then just multiply it by 180.0640

So then, 180 times 18...0...this is 64...8 + 6 is 14...then put the 0 here; 0, 8, 1; 0, 4, 12...it is going to be 3240 degrees.0649

All 20 angles of a 20-gon are going to add up to 3240 degrees.0684

Looking at the formula, it is going to be the number of sides; subtract 2 (you are going to solve this out first)0695

to figure out how many triangles you have in that polygon; and then just multiply it by 180; that is it.0705

It is just the number of sides, minus 2: take that number and multiply it by 180, and that is going to give you the interior angle sum.0712

Now, the exterior angle sum theorem ("exterior" meaning outside): whatever you have...it can only be one exterior angle0726

from each side or vertex...then if this right here is 1, this is 2, this is 3, this is 4, and this is 50742

(there have to be 5 of them, because there are only 5 sides here; it is a pentagon)--all 5 angles here are going to add up to 360.0757

And that is the exterior angle sum theorem; the interior angle sum theorem is different,0767

because depending on the number of angles, depending on what the polygon is, the interior angle sum is going to be different.0771

The more angles the polygon has, the greater the sum is going to be.0779

But the exterior angle sum is always going to be 360--always, always, no matter what type of polygon you have,0785

whether you have a triangle (if you have a triangle, it doesn't matter if you measure the exterior angle this way,0795

as long as you do the same for each vertex: let's say 1, 2, 3 here; the measure of angle 1, plus...they are all going to add up to 360)...0802

Here, the measure of angle 1 + 2 + 3 + 4 + 5 are all going to add up to 360; and that is the exterior angle sum theorem.0821

The first example: Draw two figures that are polygons and two that are not.0832

You can just draw any type of figures that you want.0838

The first two that are polygons...you can just draw...it doesn't matter...any type of polygon.0844

You can draw something like that, as long as it is closed, each side is a straight line segment, and no sides are overlapping.0854

Then, two examples, two figures, that are not polygons would be exactly those things.0874

Maybe like that...that would not be a polygon; if I have two sides crossing like that, that is a non-polygon.0882

If I have, I don't know, something like this, that wouldn't be a polygon...use any examples that are something like this.0895

These are polygons; these are not polygons.0911

Moving on to the next example: Find the sum of the measures of the interior angles of each convex polygon.0916

The first one is a heptagon: now, a heptagon is a 7-sided polygon; this has 7 sides.0924

Remember: if the number of sides is 7 (n is 7), we have to figure out how many triangles.0933

Remember: you subtract 2, so the number of triangles is going to be 5; and then, you are going to multiply that by 180.0940

Now, the formula itself is going to be that the sum is equal to 180 times (n - 2).0957

This is 7 - 2; that is 5; so it is the same thing as 180 times 5.0971

Then, you can just do it on your calculator; I have a calculator here; 180 times 5 is going to be 900, so the sum is 900 for a heptagon.0978

The next one is a 28-gon; now, once you pass 12-sided polygons, there is no name for it, so you would just write 28 with "-gon."0995

This is a polygon with 28 sides, so n is 28; the number of triangles is 26; you are going to multiply that by 180.1006

To use the formula, you are going to do 180 times number of sides; that is 28 - 2, so 180 times 26...use your calculator...is going to be 4680 degrees.1021

So, all the interior angles of a 28-gon are going to add up to 4680 degrees.1051

And the next one, x-gon: now, for this one, we don't know how many sides there are in this polygon.1061

So, you are just going to use the formula; and so, we know that n is going to be x; the number of sides is x.1067

In the formula, you are just going to replace the n with the x.1079

It is supposed to be the number of sides, minus 2; instead, we are going to say x - 2.1083

And that would be it; you are just replacing the n with whatever they give you as the number of sides, and that would be x.1088

Given the measure of an exterior angle, find the number of sides of the polygon.1104

Before we start with these numbers, with these examples here, I want to first use a triangle.1112

Now, we know, from the exterior angle theorem, that the sum of the exterior angles is always 360.1120

The sum of the measures of all of the exterior angles is going to be 360.1127

If I have a triangle, here, here, and here: those are my three exterior angles: 1, 2, 3.1134

How would I be able to find the number of sides?1146

Well, in this case, how would I find each of these angle measures?1153

Wouldn't I have to do 360 degrees, divided by the number of exterior angles?1161

This is going to be what?--each of these angles has to be 120.1172

Now, again, this is going to be for a regular polygon; for a regular polygon, this is going to be 120;1181

if all of these exterior angles have the same measure, then it will be 120 each.1188

So, that way, it will total to 360.1194

Well, it is like they are giving you the measure of each exterior angle; so how would we figure it out...1197

if I said that each exterior angle has an angle measure of 120--each exterior angle of a polygon is 120--find the number of sides.1205

Well, you would have to do the same thing: 360 (because that is the total) divided by 120, and that is going to give you 3.1218

So, you know that there are three sides here.1227

The same thing works for this: 36 is the angle measure of each exterior angle.1233

So, if 360 degrees is the sum of all of the exterior angles, divide it by 36 to find the number of sides; you get 10.1241

That means that the polygon has 10 sides.1256

The same thing works here: 360, divided by 15 degrees (you can use your calculator), is going to give you 24 sides.1266

If there are 24 sides, each angle of the sides (because if there are 24 sides, that means that there are 24 angles)--1293

each exterior angle--has a measure of 15 degrees, which will then1303

(since there are 24 of them) add up to 360 when you multiply these two together.1308

The same thing works for this one, x: each exterior angle measure is going to be x degrees--we just have to divide it.1314

And since we can't solve that out, this will just be the answer; you are just simplifying it out as much as possible, and that will be it.1328

There is nothing else that you can do with that.1337

The fourth example: Find the sum of the interior angles of each polygon.1342

Now, notice how both of these polygons are not regular polygons; it doesn't look like it is equilateral; it doesn't look like it is equiangular.1348

But it is OK, because we are just looking for the sum of all of the interior angles.1358

Since it is not regular, we would not be able to find the measure of each angle.1367

But instead, we can find the sum--what all of them add up to--because it depends on the polygon, not the type of angles inside the polygon.1374

Here, we have 1, 2, 3, 4, 5, 6; so we have a hexagon, a 6-sided polygon.1383

And that means, in a 6-sided polygon, that we have 4 triangles.1391

Remember: we subtract 2; we get 4 triangles; and then we have to multiply this by 180.1399

4 times 180...this is 32...720; that means that the sum of all of the angles inside here is going to be 720.1409

Now, if, let's say (I am just going to add to this problem here), this was a regular polygon--1431

say that all of the sides are the same, and all of the angles are the same, so it is equilateral, and it is equiangular;1443

and I want to find what the measure of each angle is, then.1450

Since each of these angles are the same, and I know that all 6 angles together1453

are going to add up to 720, how can I find the measure of just one of them?1460

Since they are all the same, how can I find the measure of just this one, the measure of angle A, or the measure of 1?1466

Since they all have the 720, and they are all the same--they all have the same measure, and there are 6 of them,1478

I can just take 720 and divide it by 6; 720/6 is going to give me the measure of each of these angles.1484

So then, here, you do 720 divided by 6; each of these angles is going to be 120; 120 here, 120 here, here, here, here, and here.1499

And that is only if you have a regular polygon, meaning that all of the angles are the same.1521

All of the angles have to be the same for you to be able to divide your angle sum to figure out each of these angle measures.1527

The next one: here, this is to find the sum of all of the angles side.1539

This is a quadrilateral; we only have four angles; so this is just going to be 180 times 2: let's just use the formula...1545

180...n - 2 is the sum; 180...we have four sides, minus the 2, so that means we have two triangles;1554

180 times 2, we know, is 360 (I said 360, and I wrote 320).1569

Now, again, if all of these angles were the same, were congruent, this is equilateral and equiangular, so it is a regular polygon.1579

Then, you would take 360; you can divide it by 4; and that would just be 90 degrees; that is if each of these angles were the same.1593

Then, each of them would have a measure of 90; and we know that that would just make this a square, if it was an equilateral, equiangular quadrilateral.1606

That would make it a square; then you would know that each of these angles would have to be a right angle.1614

But for the sake of just knowing what to do if you have a polygon that is regular--1618

not just a quadrilateral, but any other type of regular polygon--you would just take the sum,1624

and divide it by the number of angles you have.1631

And that is it for this lesson; thank you for watching Educator.com.1641

Welcome back to Educator.com.0000

For the next lesson, we are going to go over area of parallelograms.0002

Now, a parallelogram, remember, is a polygon with two pairs of parallel sides.0007

So, these two sides are parallel, and these two sides are parallel.0013

To find the area, we are going to do the base times the height; so it is the same formula as a rectangle or a square.0019

But just remember that, although this is the base, the height has to be the length that is perpendicular to the base.0029

So, it doesn't actually matter which side you label as the base; this can be the base; this can be the base; this can be the base.0040

It does not matter, as long as the height is the length perpendicular to that base.0051

If I am going to call this the base, then the height has to be from here to here, so that it is perpendicular.0061

This would be the height; be careful not to make this the height.0077

If this is the base, then this is not the height; if you were to measure how tall you were, your height, you would stand up straight.0084

You would stand perpendicular to the ground to measure your height, just like this.0094

You are not going to stand to the side; you are not going to bend over and then measure you height from there.0101

You have to stand up straight--that is perpendicular to the ground.0109

That is the area: it is the base times the height; and again, the height has to be perpendicular.0113

To find an area of a figure that is not a parallelogram or a rectangle or any of the types of polygons that you are used to,0123

then you are going to have to break it up into parts.0134

So here, these lines are drawn for you; but if you were to just have this figure here, and it said "find the area,"0137

then you can break it up into parts: you can break it up into here and here; then you would have to find the area0151

of this rectangle here, find the area of this rectangle here,0157

and then find the area of this triangle here, and then we are going to add them all up--it is this plus this plus this.0161

Or you can maybe cut it down here; this would be one big rectangle; and then, this can be a trapezoid.0168

Now, we are going to go over trapezoids in the next lesson; but if you know the formula to the trapezoid,0178

you can add up this rectangle and the area of that trapezoid together, too.0184

That would be the area of figures; now, let's go over our examples.0192

Find the area of the shaded area, or shaded region: that would be everything in blue.0197

That means that you would have to find the area of this whole thing (that is the area of this parallelogram here), and subtract the area of this rectangle.0206

Now, I am going to tell you that is not good to assume that this is a parallelogram and this is a rectangle.0220

But I am just going to tell you that it is: this is a parallelogram, so show that it is a parallelogram, like this: this side with this side.0229

And then, I am going to draw that to show that it is a rectangle.0240

Again, take the area of the parallelogram, and then subtract the area of the rectangle.0248

It is as if I have a full parallelogram, and then you are going to cut out this rectangle; don't you have to subtract it and take it away?0255

The area of the parallelogram is going to be base times height; the base is 12;0266

the height is not 10--it has to be the perpendicular one, so it is going to be 9.0281

12 times 9...that is going to be 108; and then, you are going to subtract the rectangle.0287

The area of a rectangle is going to be base times height here; so that is 8 times 5; that is going to be 40.0302

And then again, you are going to take the area of the parallelogram, minus the area of the rectangle, and that is going to be the area of the shaded region.0319

So then, it is going to be 108...subtract the 40...and that is going to give you 68; that would be units squared,0332

because any time you are looking at area, it is always units squared.0348

The next example: Find the height and the area of the parallelogram.0362

We have to first look for the height; they give us the base; they give us this length; and then they give us this angle measure.0368

Well, if this is 45 degrees, this is a right triangle; using this right triangle, I can find this measure right here.0379

Here (remember special right triangles?), if you have a 45-45-90 degree triangle0393

(if this is 45, this has to be 45, so this is a special right triangle: 45-45-90), if this is n, the measure of the side opposite the 45-degree angle--0402

so then, if this is 45, then the side opposite this angle is going to be this side right here;0416

and the side opposite this angle is that side; so this angle and this side go together.0423

This is 45-degree; this is n, with that side opposite; then the side opposite this 45 is also going to be n, because they are the same; it is isosceles.0432

Then, the side opposite the 90, which is the hypotenuse, is going to be n times √2.0444

Now, that is this side right here, which is 10; so they give you this side--this is 10.0453

That means that n√2 is equal to 10; so, I can just make n√2 equal to 10.0465

Do you remember this section from special right triangles?0482

A 45-45-90 degree triangle is going to be n, n, n√2; so I am just going to make these two equal to each other: n√2 is equal to 10.0486

Then, I am going to solve for n, because this is what I want to solve for.0498

I want this measure here, because that is going to be the height.0502

Even though this is n for now, this is also h for height.0507

So, if I solve for n here, I need to divide the √2; so n = 10/√2.0511

Remember: I have to rationalize this denominator, so I have to multiply this by √2/√2.0521

This becomes 10√2/2; if I divide this, I am going to get 5√2, so n is equal to 5√2; this is 5√2.0528

Since that is also the height, I can just go ahead and say that my height is 5√2.0545

So, I have my base, and I have my height; now I can solve for the area.0556

Area equals 12 times 5√2; this is going to be 60√2 units squared.0561

Or, if you want, you can just punch it in your calculator; I have my calculator here, and that will be 84.85 units squared.0580

Example 3: Find the area of the parallelogram, given the coordinates of the vertices.0612

For coordinates, since these are the sides of the parallelogram, we want to know what we can name as the base--0621

what we can measure as the base, the length, the distance between the points--as the base and the height.0634

Remember: the base and the height have to be perpendicular to each other.0641

So, we have to make sure that the distance between these points is going to be perpendicular,0643

which means that their slopes have to be negative reciprocals.0649

If you are a visual person, you can go ahead and graph these points out, just to help you visualize what the parallelogram is going to look like.0657

If I have (10,-4), that is going to be in quadrant 4; so, this is 10; this is -4.0673

And then, (4,-4)...let's say this is (4,-4)...is right there; (4,-2)...this is (4,-2)...is right there; and (10,-2) is right there.0687

We know that this is going to be a rectangle; how do I know?--because these two points have the same y-coordinate;0705

these two points have the same x-coordinate; and the same here--these have the same x-coordinate; this has the same y-coordinate.0715

So, these two are horizontal, and these two are vertical.0722

Remember that rectangles are a special type of parallelograms, because with rectangles, the opposite sides are parallel.0730

So then, that is also a parallelogram.0739

Now, we know that, in a rectangle, these sides are perpendicular to each other; so we don't even have to worry about finding slope.0743

If you need to, then you would have to find slope; if you remember, the slope, m, is (y2 - y1)/(x2 - x1).0753

We don't have to do it for this one; but if you have a problem like this,0766

where you have to find slope to see if they are actually perpendicular,0770

then just take the y-coordinate, subtract it from the other y-coordinate, and divide it by the difference of the two x-coordinates.0774

That is going to give you slope; and then, you would have to do that for all of the lines to see if they are negative reciprocals of each other,0785

meaning that, if the slope of two points was 2, then the negative reciprocal is going to be -1/2.0793

It is the negative, and then the reciprocal; so if this is 2/1, then it has to be 1/2,0807

and then that would mean that this slope and this slope are perpendicular to each other.0812

Here, since we know that they are perpendicular, we can just call this the base and call this the height.0823

To find the area, I know that this base is going to be 6, because it is 6 units away, from 4 to 10; the base is 6;0834

and then, the height...this is -2, so the measure from here to here is 2; so the area is going to be 12 units squared.0846

And the fourth example: we are going to find the area of this figure.0872

We need to break this up into several parts; you can do this different ways.0878

You can cut it here; you can cut it here if you want; you can cut it here, and then cut it here, if you like.0886

It doesn't matter; let's see, what do I want to do? Let's just do it this way.0892

I am going to cut it here (so then, we have this rectangle), and then cut it here (so we have this rectangle).0900

Now, be careful; since I cut this up, I can't use this for the base; I have to use this as the base.0907

So, here, this area of this is going to be 7 times 10 (the base times the height--make sure that it is perpendicular;0916

if you have a parallelogram, make sure you have the height); so this would be 70.0927

Then, this: do I know what the base is here?0934

Well, if this whole thing is 10, and this one is 7, then this has to be 3.0939

And let's see, I don't know what this is; well, if this is 3, this is 3, and from here all the way to here is 100951

(because that is what this number tells me), this is 3, this is 3, and that is 6; that means that this has to be 4,0966

because this plus this plus all the way down here has to be 10, so this is 4.0975

That means that, for this rectangle here, I am going to make 3 the base and 7 the height.0983

Make sure that you don't multiply this; this thing right here is going to be 3 + 4, so this is going to be 3 times 7, which is 21.0993

So, let me just circle it, so that you know, and you don't get confused with my numbers.1003

And then, for this rectangle here, we have 11, and then we know that this is also 4, because we found it there.1008

It is just going to be...from here to here (make sure that you don't use this number, because this number1016

is showing you from here all the way to here) is going to be 11 times 4, which is 44.1019

So now, all we have to do is add up these numbers: so 70 + 44 + 21...we have three rectangles,1029

so we are adding up 3 numbers; so 4 + 1 is 5; 7 + 4 is 11, plus 2 is 13;1041

that means the area of this figure here is going to be 135 units squared.1050

And that is it for this lesson; thank you for watching Educator.com.1063

Welcome back to Educator.com.0000

For the next lesson, we are going to go over area of triangles, rhombi, and trapezoids.0002

First, let's go over the area of a triangle; now, we have been doing this for years now: it is 1/2 base times height.0008

Now, the reason why it is half base times height: let's say I have a parallelogram.0019

A parallelogram is a quadrilateral with two pairs of opposite sides parallel.0030

Now, the area of a parallelogram, whether it be this type of parallelogram, a rectangle, or a square, is base times height.0038

To get a triangle from a parallelogram, we have to cut it in half; if we cut a parallelogram in half, we get a triangle.0051

So, we are dividing it by 2; so, the triangle is 1/2 base times height.0064

Now, base times height, divided by 2, is the exact same thing.0076

Think of a triangle as half the area of a parallelogram; a parallelogram is base times height, so it would just be base times height, divided by 2.0082

And it is important to keep in mind that if this is the base (it doesn't matter which one you label the base,0096

but it is always easiest to just label the bottom side the base), then the height has to be the length from the base0103

to the vertex opposite that base, so that it is perpendicular.0119

If you are going to name this the base, then this has to be the height; it is 1/2 the base times the height.0127

Make sure that this is not the height; height has to be straight vertically, perpendicular to the base.0135

So again, the area of a triangle is 1/2 the base times the height.0147

Next is the trapezoid; now, the trapezoid formula for area is 1/2 times the height times the two bases added together, the sum of the two bases.0153

Now, it looks a little long and complicated, but it is actually not; if you think about it, it is actually the same as the parallelogram.0169

The area equals base times height: now, it is the same formula, but the reason why it is kind of complicated0178

is because the base here...when it comes to a parallelogram, let's say a rectangle,0186

we know that, if we are going to label this the base, well, this is also the base, too; this is the base, and this is the base.0200

They are the same, so we don't have to worry about two different numbers for the base, because they are exactly the same.0208

When it comes to a rectangle, if I talk about "base," then I could be talking about this one or this one, because they are exactly the same.0217

When it comes to a trapezoid (and by the way, a trapezoid is when you have one pair of opposite sides parallel--only one),0226

well, we have two different bases; and remember, bases, in this case, have to be the parallel sides.0239

So, this would be one of the bases, base 1; and the side that is parallel, opposite, to it, will be base 2.0246

They have to be the bases; you can't call these bases--they are the legs (these are called legs).0261

But here is a base, and here is a base; now, unlike our rectangle, where these opposite sides,0268

both being bases, are exactly the same--here our bases are different.0277

So, for this formula, we would just have to look at this base again; it is the average of the two bases.0282

We are using the same exact formula, but this represents the average of the two bases, because the bases are different.0292

Now, if I rewrite this formula, I can write it as height, times base 1 plus base 2, divided by 2.0302

All I did here was to take this 1/2 and put it under the two bases, the sum of the bases, right here.0318

Now, if I do this, then how do I find the average?0327

I have to add them up and divide by the number--whatever I have.0332

So, this can be considered the average of the two bases; again, it is the same thing, base times height;0338

but then, the base wouldn't just be any base, because we have two different bases; so you have to take the average of the two bases.0348

So, area equals base, or the average of the two bases, times the height.0354

Think of it that way; that way, it is just a little bit easier to remember the formula.0365

It is the height, times the average of the bases; and that way, you don't have to think of this 1/2 in the front.0370

If you want, you can just use the same formula, this formula that is written here; but you can also just use this 1/2 to make this over 2.0378

And it would just be the average of the bases, times height; so it is still base times height, but it is just the average of the bases--0391

the two bases, added together, divided by 2.0398

And again, the height has to be perpendicular to the base.0401

So, it is base times height, but the base for a trapezoid has to be the average of these two bases.0409

Now, let's say I have this height being 3, and this base has a measure of 6, and this base has a measure of 8.0417

Again, area equals base times height; but since I have a trapezoid, I have to find the average of the bases;0434

so it is going to be 6 + 8, divided by 2, times the height, which is 3.0444

6 + 8 is 14, divided by 2 is 7; so the average of 6 and 8 is 7, so 7 is actually going to be the number that we are going to use as our base.0458

That, times the 3, is 21; so 21 units squared--that would be our area.0471

Moving on to the rhombus: now, if you only have one, it is called a rhombus; if you have more than one, the plural is rhombi.0486

Now, a rhombus is a quadrilateral (a four-sided polygon) with four congruent sides.0504

Now, these angles are not perpendicular; if they were, it would be considered a square; it is just an equilateral quadrilateral.0511

Now, with these four sides, they form two diagonals; there is one diagonal, and there is another diagonal:0523

diagonal 1 and diagonal 2--it doesn't matter which one you call diagonal 1 and which one you call diagonal 2.0538

There are two of them, and you are going to be multiplying both of them together and then dividing it by 2.0544

Now, these diagonals, for any rhombus, are going to be perpendicular; so again, 1/2 times the two diagonals...0551

you can think of it as diagonal 1, times diagonal 2, and then divided by 2; in this case, it is not the average of the diagonals,0568

because to find the average, you would have to add up the two diagonals and then divide it by 2; here we are multiplying.0575

Multiply this diagonal by this diagonal, and then divide it by 2; and that is the area of a rhombus.0582

Let's go into our examples: the first one: we are going to find the area of the polygon.0593

Now, since we see these little symbols right here, I know that these two sides are parallel.0597

That means that, since that is the only pair of parallel sides that I have, this is a trapezoid.0605

To find the area of a trapezoid, it is still going to be base times height; but because we have to different bases, we have to find the average of those bases.0611

So, to find the average, we add them up and divide by however many we have.0624

In this case, we have two bases, so we are going to do 9 + 11, divided by 2, times the height; and this is the height.0630

Let me just do that, so that you know that that is perpendicular.0644

It is going to be times 6; area equals...9 + 11 is 20; 20/2 times 6...10 times 6 is 60, and that is inches squared.0647

Remember: with area, you always have to make it units squared; and that is the answer.0668

The next example: Find the area of the figure.0680

Now, this is a 1, 2, 3, 4, 5-sided polygon, but we don't have a formula for just any five-sided polygon.0685

What you would have to do is break it up into two parts, two different polygons: we have a triangle up here, and we have a rectangle down here.0701

And then, once you find the area of this and find the area of this, we just add it together.0709

Let's see, for the rectangle...the area of the rectangle plus the area of the triangle...that is going to give us the area of the whole thing.0718

First, the area of the rectangle: well, we know that it is base times height, so that will be base times height;0736

for the triangle, remember, it is half a parallelogram; so it is just base times height divided by 2, or this.0751

And we are just going to add them all up: so here, the area of a rectangle is 10 times 12, which is going to be 120.0763

For the triangle, we have 1/2...what is the base?...well, it doesn't tell us what this is, but it tells us what that is;0777

and we know that, since this is a rectangle, this is going to also be 12.0786

So, that is 12 as the base, and the height is 8; make sure that you use the height that is perpendicular to the base.0797

This is...you can, just to make it easier on you, put this over 1; and then you can cross-cancel these.0807

So then, this is divided by 2, so it becomes 6; so that will be 6 times 8, which is 48; so the area of the rectangle,0817

plus the area of the triangle, is going to give us 168; the units are meters squared.0832

Any time you have area, you are always going to do units squared; so this is the area of this figure.0847

OK, the next example: we are going to find the area of this figure.0857

Let's see, we have here a rhombus; I know that that is a rhombus, because I have four congruent sides, and the diagonals are perpendicular.0860

So, here is a rhombus; and this is a trapezoid, because we have one pair of parallel sides.0879

I can just find the area of this, find the area of this, and then add them together.0890

So first, to find the area of the rhombus: area is 1/2 diagonal 1 times diagonal 2.0897

I multiply the diagonals together, and then divide it by 2: 1/2 times...0917

now, if this is 4, this whole diagonal...don't just consider this; this is only half of the diagonal, so this whole thing is 8;0925

and this whole diagonal...if this is 6, then this is also 6, and this whole thing is 12;0941

and we are just going to multiply it all together.0950

Now, you want to cross-cancel out one of these numbers; it is probably just easier to cross-cancel out the bigger numbers.0953

You can just make that into a 6; 8 times 6 is 48 units squared.0962

And then, for our trapezoid, area is base times the height, but remember, because we have two different bases0975

(the bases are the two parallel sides), we have to take the average of those two bases, so add them up and divide by 2.0993

Keep in mind: even though the bases, the two parallel sides, are here and here, and it might seem like,1002

(since this is the one on the bottom--this is the side that is on the lower side, the bottom side)...that is not considered the base.1011

It has to be the two parallel sides, 5 and 7; so 5 + 7, divided by 2, times the height...1017

now, here they don't give us the height of this, but we can use this;1027

now, this is supposed to be the same as this, so this will be 6.1030

5 + 7, divided by 2...my average is 6, because this is 12, divided by 2 is 6, times 6, which is 36 units squared.1044

To find the area of the whole thing, I am going to take the area of the rhombus, 48,1061

and add it to the trapezoid--that is 36; and that is going to be 84 units squared.1070

For the fourth example, the area of a trapezoid is 60 square inches, and its two bases are 5 and 7, and we are going to find the height.1091

In this case, the area is given; the measures of the bases are given; and then, we have to find the height.1103

First, let's draw a trapezoid: the area is 60...parallel, parallel; the shorter base is 5, and then 7; we want to find the height.1113

h is what we are looking for: now, remember the formula for the trapezoid.1133

It is the average of the bases, times the height; so it is base times height, but it is just the average of the bases.1140

Here, area is 60; that equals...my two bases are 5 + 7, over 2; and h is what I am going to be looking for.1149

Now, let's simplify inside the parentheses and find the average of the bases: 5 + 7 is 12, divided by 2 is 6.1167

Here, to solve for h, I am going to divide the 6; so I get 10 as my height.1186

And this is all in inches, so my height will be 10 inches.1195

If you are given the area, and you have to look for a missing side, base, height...whatever it is,1207

just plug everything into the formula and solve for the unknown variable; solve for what you are looking for.1218

Make sure that you don't forget your units.1225

And that is it for this lesson; thank you for watching Educator.com.1228

Welcome back to Educator.com.0000

For the next lesson, we are going to go over area of regular polygons and circles.0002

To review for a regular polygon, we know that it is when you have a polygon0008

with all of the sides being congruent and all of the angles being congruent; it is equilateral and equiangular (it has to be both).0015

Now, a couple more things to review for this lesson: if this is the center of my polygon,0024

I know...let's say I am going to have a starting point right here...that to go from here,0033

all the way around, we know that this is 360 degrees; to go all the way around a full circle is 360 degrees.0042

Also, for right triangles: Soh-cah-toa is for right triangles when you are given angles and sides, and you have to find an unknown measure.0055

This is "the sine of an angle is equal to the opposite side over the hypotenuse."0079

"The cosine of the angle measure is equal to...a is for adjacent, over the hypotenuse."0089

And "the tangent of the angle measure is equal to the opposite over the adjacent side."0100

That is Soh-cah-toa; and also, for right triangles, there are special right triangles.0111

We have 30-60-90 (let me do it on this side) triangles; the side opposite the 30-degree angle...let's say this is 30,0118

and this is 60...is going to be n; the side opposite the 60...be careful here; it is not 2n; it is n√3;0136

and the side opposite the 90 is going to be 2n.0152

That is a 30-60-90 special right triangle; and also, a special right triangle is 45-45-90.0156

This is 45; this is 45; this is 90; the side opposite the 45...if that is n, then this is also n, because they are going to be congruent.0170

The side opposite the 90 is going to be n√2; so here is the second type of special right triangles.0184

So again, when you are going all the way around a full circle, that is 360 degrees.0193

For right triangles, you can either have a special right triangle--if you have a 30-60-90 or a 45-45-90--0201

then you can use these shortcuts; if not, then you would have to use Soh-cah-toa.0209

The formula for the area of a regular polygon is 1/2 times the perimeter, times the apothem.0222

Here is a new word, apothem, and we are going to talk about that in a second.0233

To explain this formula, imagine if I have my regular polygon, because this is the area of a regular polygon (I'll draw that a little better).0239

If I take my polygon, and let's say I break it up into triangles; from the center, I am going to create0255

a triangle here, a triangle here, here, here, here, and here; I know that the area of one triangle is 1/2 base times height.0269

So, if that is one triangle, 1/2 base times height...how many triangles do I have?...I have 1; I have 2, 3, 4, 5, 6;0295

so, it is 1/2 base times height, times 6; and this would be the area of this regular polygon.0310

Now, of course, that is only if it is a regular polygon; you can only do this if you have a regular polygon.0321

1/2 base times height is going to give you the area of one triangle, and you are going to multiply it by the 6 triangles that you have.0328

Well, look at how many sides I have for this polygon: this is 1, 2, 3, 4, 5, 6--this is a hexagon--this is 6 sides.0335

So, right here, my base, times the 6, is going to give me perimeter, because, if this is the base,0347

I have 6 of them together; base and 6 together are going to give me the perimeter.0366

The height is this right here; this height is now called the apothem.0378

The apothem has to go from the center to the midpoint of one of the sides; that is called the apothem.0386

So now, my h is going to turn into an a; the height of one of the triangles is now an apothem.0395

And then, 1/2 is going to be part of the formula, as well.0403

Just changing my base times the 6 to become perimeter, and then renaming the height to be the apothem0413

(the height of the triangle is an apothem of the regular polygon), and then just keeping the 1/2...0426

that now becomes the formula for the area of a regular polygon.0433

If you ever get confused with this formula right here, all you have to do is just break this up into triangles;0439

find the area of one of the triangles, and multiply it by however many triangles you have.0448

And that is going to be the exact same thing as this formula here.0451

But this is the main formula for a regular polygon: 1/2 times the perimeter times the apothem.0457

The area of a circle, we know, is πr2.0477

From the center to a point on the circle is called the radius, and that is what this r is--the radius.0481

The radius squared, times π, is going to give you the area of this whole circle.0494

OK, let's work on some examples: Find the area of the regular polygon.0503

Now, to find the area of this polygon, my formula is 1/2 times the perimeter, times the apothem.0511

Now, I have that this side is 10; and again, this is a regular polygon, so then we know that all of the sides have to be 10.0526

I can find the perimeter: a = 1/2...the perimeter would be 10 times however many sides I have; that is 5, so it is 10 times the 5.0532

And then, for my apothem, remember: the apothem is from the center to the midpoint of the side, and it is going to be perpendicular.0547

And if it is the midpoint, well, we know that this whole thing is 10--this whole side has a measure of 10.0564

Each of these is going to be 5: 5 and 5.0571

But I need this right here, a; that is my apothem--that is what I need.0578

What I can do is draw a triangle from here to there; we know that it is a right triangle; and I can look for my side.0585

But for me to do that, I need to know this angle measure right here.0600

What I can do is just...let's say I have all of these triangles, again; I know that from here, all the way around here--that is a total of 360 degrees.0605

A full circle is 360 degrees; now, I just want to know this little angle measure right here.0628

What I can do: since it is a regular polygon, I can take 360 degrees, and I am going to divide it by however many triangles I have here.0637

I have 1, 2, 3, 4, 5--5 angles that make up my five triangles; so 360 divided by 5 (I have my calculator here) is going to give you 72 degrees.0652

That means that, for each of these, this is 72; this is 72; 72; 72; and this whole thing right here is going to be 72.0678

But since I am only using this half-triangle, because I want to make it a right triangle0695

(because if I use a right triangle, then I have a lot of tools to work with; I have a lot of different things that I can use;0704

I can use Soh-cah-toa; I can use special right triangles), I want to use this right triangle here;0712

so then, I am going to take this 72, and then divide it by 2 again, because I just want this angle measure right there.0722

So, 72 divided by 2, again, is going to be 36.0729

That means that this angle right here, where the arrow is pointing, is 76; that angle is 76.0738

Now, I am going to re-draw that triangle, so that it is a little bit easier to see; that is my apothem; this angle measure is 36 degrees;0746

this side right here is 5; now, since this angle measure is 36, I can't use my special right triangles,0762

because only when it is a 30-60-90 triangle or a 45-45-90 degree triangle could I use special right triangles.0773

Since I can't use special right triangles, I would have to use Soh-cah-toa.0780

Now, if you remember, Soh-cah-toa is made up of three formulas.0786

I just have to figure out which one I am going to use: I am not going to use all three--I am just going to use one of these, depending on what I have.0796

Look at it from this angle's point of view: I have the side opposite, so I have an o; and I have the side adjacent, so I have an a.0804

Then, which one uses o and a? This uses h; this uses ah; this one uses oa, so then I would have to use tangent.0820

That is going to be...the tangent of this angle measure, 36, is equal to the side opposite, which is 5, over the side adjacent, which is a.0831

Now, I can just solve for a; so I am going to multiply this side by a and multiply the other side by a.0851

That way, this is going to be a times the tangent of 36, this whole thing, equal to 5.0860

Remember: keep these numbers together--it has to be the tangent of an angle measure; you can't divide the 36.0870

You have to find the tangent of 36 on your calculator.0876

So, to find a, I am going to, on the calculator, do 5 divided by this whole thing, tan(36): with 5/tan(36), I get my as 6.88.0880

My apothem is 6.88; so now, I have my apothem--this is going to be 6.88.0919

Then, from here, I just have to solve this out: this is 1/2, times this whole thing (is the perimeter), times the apothem.0934

You could just do that on your calculator; and so, the area should be 172.05 units squared0946

(because it is still area, so make sure that you have your units squared); I'll box it to show that that is my answer.0970

So again, since we have a side, we can find the perimeter; the perimeter is fine.0981

But to find the apothem, you are going to have to take the 360, divided by however many sections that you need it to divide into.0989

So, that way, you can use this right triangle right here, and then you just use Soh-cah-toa to find the apothem.0998

All right, let's do a couple more: the next example: Find the area of the regular polygon.1007

Here, this is a; even though it is not going down to this side, it is still an apothem--we know that this is the apothem; it is 5.1016

And that is all they give us; remember: to find the area of a regular polygon, it is going to be 1/2 times the perimeter times the apothem.1032

We have the apothem, but no sides, so we can't find the perimeter.1043

So, what we can do is, again, form this triangle here; that way, we have a right triangle.1049

And then, since I need an angle measure (because if I find this side right here, then I can multiply it by 2,1058

and then that is going to give me the whole side), again, break this up into triangles--just sections--1069

so that you know what to divide your 360 by.1080

From here, going all the way around, we have 1, 2, 3, 4, 5, 6, 7, 8, which is the same as the number of sides that we have; this is an octagon.1087

So, it is 360, divided by 8; that is going to give us the angle measure of each triangle: 360/8...we get 45.1104

But remember: you have to be very careful, because that is for each of these triangles right here.1125

So, this whole thing right here is 45; each of these angles is 45, and so is this right here.1131

But again, we need to find only this angle measure, which is half of the 45,1145

because the 45 is the same as from here all the way to here; that is 45.1151

So, we are going to take the 45 and divide it by 2; and then, that angle measure right here is going to be 22.5.1159

I am going to re-draw this right triangle; this is my apothem that was 5; this angle measure is 22.5 degrees.1169

And I am looking for this side, so then, let's label that x.1187

I have a right triangle; we need to use Soh-cah-toa; so again, from this angle's point of view, what do I have?1193

I have opposite, and I have adjacent; this is the hypotenuse, so I have opposite and adjacent, so I am going to have to use "toa."1209

That means that the tangent of the angle measure, 22.5 degrees, equals...the side opposite is x, over...adjacent is 5.1221

To solve for x, I am going to multiply both sides by 5; and again, use your calculator: 22.5...the tangent of that...multiply it by 5.1235

Make sure you find the tangent of this number; don't multiply this times this and then find the tangent of that--that is going to be wrong.1257

It is 5 times the tangent of this number; and that is going to give me 2.07.1265

So, this right here is 2.07; but remember: I need this whole thing--that is my side.1275

If just this is 2.07, then the whole thing is times 2; so, my side, then (I can just write it here, because this side is the same thing as this side) is 4.14.1287

So then, go back to our formula: area equals 1/2...the perimeter would be 4.14 times...how many sides do we have?1314

This is an octagon, so we have 8 sides; so then, this together would make the perimeter...times the apothem.1330

The apothem is 5; that was given to us; so, my answer, then...what you can do is just punch in this times this times this,1339

and then just divide it by 2, because multiplying by 1/2 is the same thing as taking all of that and dividing it by 2.1358

The area is going to be 82.84 units squared.1367

Again, just to review what I just did: the apothem was given to me, but that was the only measure that I had.1381

So, I had to take the 360 and divide it by 8 to find the angle measures for each of the triangles.1390

But then, I cut that triangle, again, in half, to make it a right triangle; so then, this angle measure right here was 22.5.1401

I used Soh-cah-toa: the tangent of 22.5 equals x over 5; you solve for x, and that is going to give you...1412

remember this section right here, this little part right here; multiply that by 2 to get the full side.1424

And then, from there, since I have one of the sides, and this is equilateral, I just multiply that by 8, because there are 8 sides total.1433

That gives me the perimeter; the perimeter, times the apothem (which is 5), multiplied all together...you get 82.84 units squared.1441

The next one: Find the area of the shaded region.1453

Here we have a circle; and inscribed in that circle is a hexagon, because we have a 6-sided polygon.1458

OK, now, that means...to find the area of a shaded region, I am going to find the area of the circle.1469

It is the circle, minus the hexagon; and that is going to give us this shaded part.1485

It is like we are cutting out that hexagon from the circle.1503

To find the area of the circle, we need radius; the area of a circle, we know, is πr2--that is the area.1509

r is the radius, so it is from the center; now, ignore the hexagon for now--just look at the circle.1526

It is from the center, and it goes to a point on the circle, so that would make that the radius; it equals π(8)2.1534

So, 8 squared is 64; you are going to do 64 times the π, and that will be 201.06 units squared.1549

And now, to find the area of the hexagon: the formula is 1/2 perimeter times apothem, and this is a regular hexagon.1574

Now, be careful here: this is not my apothem; the apothem has to go from the center, and it has to go to the midpoint of one of the sides.1596

I don't have the apothem, so I need to solve for it.1613

So again, I need to take my 360 (because that is a full circle), and I am going to divide it by...1618

if I cut this up into triangles, it is going to be 6 triangles; remember: it is the same number of sides that I have.1632

So, 360, divided by 6, is going to give me 60 degrees; so that means that this whole thing right here is going to be 60.1640

But since I only want to know this angle right here, this is going to be half of that, which is 30.1656

So again, let me just draw out that triangle: here is a; this is 30; and this is 8.1664

Now, if this is 30, then this has to be 60, because all three have to add up to 180; or these two have to add up to 90, since this is a right angle.1676

30-60-90--it is my special right triangle; so, I don't have to use Soh-cah-toa--I can just use my special right triangle shortcut.1687

If this is n, then the side opposite the 60 is going to be n√3; the side opposite the 90 is going to be 2n.1702

That is my rule for a 30-60-90 triangle.1712

What is given to me--what do I have?1716

I have the side that is opposite the 90, so I have a 2n; I have this right here.1719

I want to solve for the side opposite the 60, because that is the apothem.1727

Well, 2n is the same thing as 8, so I am going to make them equal to each other.1734

How do I solve for n? Divide the 2; n is 4.1740

That is this right here: 4. Then, the side opposite my 60 is going to be n√3, so that is 4√3: a, my apothem, is 4√3.1746

Now, do I know my side? I don't have my side measures, but since I have this triangle, I can just look for this right there, which is that right there.1767

Let me just highlight that: this is the same thing as that--that is the side opposite the 30, which is 4; that means that this is 4.1780

So, this whole thing right here, this side, is going to be 4 + 4, which is 8.1795

Each of these sides has a measure of 8; so, my perimeter is going to be 8 times 6; 8 times 6 is 48; my apothem is 4√3.1803

If you want to, you can just divide this by 2, just to make it a smaller number; that is 24; multiply it by 4√3.1830

Now, since we have a decimal here, you can just go ahead and turn this into a decimal, also, by just punching it into the calculator.1842

Here, you can do 4 times √3, which is 6.93; multiply it by 24; so the area of this hexagon is 166.28 units squared.1852

Now that I found the area of the circle and found the area of this hexagon, I am going to subtract it.1885

201.06 - 166.28 is going to give me 34.78 (if you round it to the nearest hundredth) units squared.1896

So, the area of the shaded region is 34.78 hundredths.1931

And the fourth example: we have the circle inscribed in a decagon; a decagon is a 10-sided polygon.1946

And again, this is regular decagon; we are dealing with regular polygons, so this is going to be regular; find the area of the shaded region.1959

This is like the opposite of what we just did: we are taking the area of the polygon, the decagon, and we are subtracting the area of the circle.1967

Let's see, this is a decagon; let's find the area first: the area of a regular polygon is 1/2 perimeter times apothem.1978

And in this case, it looks like we are given...1995

And before you start, just look to see what you have, what you need, and how you are going to find what you need.2000

Can I find the perimeter? Well, if this side is 4, can I find the perimeter of my decagon?2011

Yes, because, if this is 4, all of the sides will be 4; so I can just do 4 times however many sides I have, which is 10.2018

That is going to give me perimeter; so, area equals 1/2 times the 4 times a 10, which is the perimeter.2026

And the apothem, remember, is the segment from the center all the way to the midpoint of one of the sides, so that it is perpendicular.2035

And that is 6; so I have everything that I need--I don't have to solve anything out.2048

Here, this is going to be, let's see...1/2 times 4 times 10, so the perimeter is 40 times this 6.2058

You can just divide one of these numbers by 2--you could cut it in half and cross-cancel.2079

So then, if I take this and a 3, I know that 4 times 3 is 12; add the 0 to the end of that number;2085

the area is 120 units squared--that is the area of the decagon.2095

I am not done: I have to find the area of the circle now.2102

The circle's area is πr2; r, which is the radius, from the center to the point on the circle--2105

well, that is going to be the same number as the apothem, because this is also, for this diagram2124

(not always, but the apothem for this diagram) the radius, because it is going from the center of the circle to a point on the circle.2130

So, that is going to be 6 squared; so that is 3.14 times 36, which is 113.10 units squared.2138

And from here, I have to take the decagon, and it is like I am cutting up the circle; so I have to subtract.2163

So, it is going to be 120 - 113, and that is going to give me 6.9 units squared.2170

So, that is the area of the shaded region here.2192

That is it for this lesson; thank you for watching Educator.com.2199

Welcome back to Educator.com.0000

For the next lesson, we are going to go over perimeter and area of similar figures.0002

If you remember from similar polygons, they have a ratio, a scale factor.0010

A scale factor is the same thing as ratio of the corresponding parts, a:b.0019

Now, if the scale factor is a:b...so let's say, for example, that this is 2, and the corresponding side for this triangle is 3--0027

again, they are similar...then the ratio, the scale factor between them, is going to be 2:3.0040

Well, then the perimeter of this first one: if the scale factor is 2:3,0049

then the scale factor of the perimeter of the first one to the second one is also going to be...0055

the ratio, not the actual perimeter, but the ratio of the perimeters is going to be 2:3; it is going to be the same.0066

For example, if the perimeter of this is 5, well, we can turn this into a fraction; so 2:3 is going to be 2/3, like that;0075

since the ratio of the corresponding parts is the same as the ratio of the perimeters,0095

I can just make it equal to 5/P, to find the perimeter of this.0106

That way, I can just cross-multiply here; if we just make this equal to P, and leave that as P, 2 times P is 2P; that equals 3 times 15.0119

I can divide the 2, and then the perimeter is going to be 15/2, which is 7.5, 7 and 1/2.0136

So, if the perimeter is 5 here, then the perimeter of this has to be 7.5.0146

Again, the ratio is going to be the same; the scale factor of the corresponding parts of this side to this side0151

is going to be the same exact scale factor of the perimeters.0157

Now, for area, it is a little bit different: if the scale factor of this triangle to this triangle is a:b, then the area of the two figures--0166

the scale factor of the area--is going to be a2:b2.0180

If this is a:b, if they are similar, then of course, the scale factor, the ratio, is going to be a:b.0188

Well, then, for this, if the scale factor of the area to the area...the area for the first one of triangle 1, let's say,0197

to the area of triangle 2, is going to be a2:b2.0216

Now, that is not actually saying that that is going to be the actual area; just because you have a:b,0226

if you square those numbers, that doesn't mean that that is going to be the actual area for the triangles.0234

It is saying that the ratio between the two areas is going to be a2:b2.0239

Let's say that a is this side right here; it is 2, and this side is 3.0247

So, the scale factor between these two triangles is going to be 2:3; that means that the scale factor0252

of the areas between this one and this one is going to be 22:32, so it is going to be 4:9.0262

Now, it does not mean that the area of this triangle is going to be 4; it is saying that the ratio of the area from this one to this one is going to be 4:9.0276

So, if the area of this is 16 units squared, then how can I find the area of this one?0286

Let's say that the area of this is what we are looking for.0302

Since I know that the ratio of the area from this one to this one is going to be 4:9, I can just create a proportion.0308

So, 4:9 is going to equal 16 (because this top number is representing this triangle; this is representing this triangle) over x.0315

We are going to label that x; then you can cross-multiply.0329

Or, since we know that 4 is a factor of 16, to get from 4 to 16, I can just multiply this by 4, which means that to get x, I can just multiply this by 4.0335

So, this will be 36; so my area here is going to be 36 units squared.0353

Now, let's just go over some examples: The ratio of the corresponding side lengths is 4:7.0369

If this one is a:b, the ratio of the perimeter is also going to be a:b; the ratio of the areas, then, is going to be a2:b2.0380

So, back to the first one: 4:7; the ratio of the perimeters is going to be 4:7.0399

Now again, that does not mean that the perimeter is going to be 4 units, and the perimeter of the second one is going to be 7 units.0407

It just means that when you simplify it, it is going to have a ratio of 4:7.0418

And then, the ratio of the areas is going to be a2:b2; be careful not to multiply it by 2--you have to square it.0424

So, 42 is 16; and 72 is 49; so this is going to be the ratio of the areas.0433

Again, it does not mean that these are going to be the areas; it just means that, when the areas are simplified, it is going to have the scale factor of 1009.0444

OK, and then here, for the second one, they give us the ratio of the perimeters.0456

This is a:b; this is also a:b; so this is going to stay at 3:2.0462

Then, the ratio of the areas is going to be 32 to 22; that is 9:4.0469

And the third one: they give us the ratio of the areas, so since this is a2 to b2, I have to take the square root,0480

do the opposite of squaring (that is taking the square root of each of these).0490

If I take the square root of this, I am going to get 13, because 132 is 169.0497

And 144...the square root of that is 12; 122 = 144.0504

Then, the ratio of the corresponding side lengths is also going to be 0792.0511

And then, for the last one, here is the ratio of the perimeters; it stays 9:10.0520

And then, the ratio of the areas...square each of those...is going to be 81:100.0526

Here, they ask for the ratios of the perimeter and the area of the similar figures.0538

Here, we have a rectangle; so if this is 6, I know that this also has to be 6.0546

And here, also, if this is 2, then this also has to be 2.0554

And I know that this side with this side is corresponding; so the ratio is going to be 6:2.0561

But then, I have to simplify: that is going to be 3:1--here is the ratio of the corresponding parts.0572

For the perimeter, the ratio is also going to be 3:1.0583

And then, the area is going to be 32, which is 9, and 12, which stays 1.0599

Now, all they wanted is the ratio of the perimeter and the ratio of the area.0614

But here, this area is given; it is 24 inches squared; so what you can do...since we know the ratio of the areas0621

(this is 9:1), the actual area for this one is given; so we can use that to look for and find this area here.0631

So again, 9:1 is going to be 9/1; change that so that, that way, we can make equivalent ratios, and that will be a proportion.0646

The area of this one is 24, and then the area of this is going to be x.0659

So here, we can cross-multiply; this is going to be 9x = 24; if we divide the 9 from both sides, then I am going to get...0668

and here, you can just simplify; this is going to be 8/3.0685

You can change this to a mixed number if you like; so then, this is going to be 2 and 2/3.0693

The area of this is going to be 2 and 2/3 inches squared.0707

The next example: Find the unknown area.0726

We have the area of this, but we don't have the area of this, so this is the unknown area.0729

Here, this is corresponding with this; so the ratio between these two figures is going to be 6:8, which simplifies to 3:4.0736

So, the ratio of this to this is 3:4; now, the ratio of the areas (I am going to write the areas separately from that)...0751

this is a:b; the ratio of this area to this area is a2 to b2; that is 9:16--that is the ratio of the areas.0764

The actual area is 54 here, and I need to find this right here; so this is going to be, let's say, x.0783

I am going to make this into a proportion: 9/16, or 9:16, equals 54:x.0794

You can cross-multiply; you can also...if this is a factor of this number, then to get from 9 to 54, you multiply by 6.0806

So, to get from 16 to x, you can just multiply by 6; and let's see, 16 (I have a calculator here) times 6 equals 96.0819

So x, this measure right here, is going to be 96; the area is 96 meters squared.0829

Again, we found the ratio of the areas; it is going to be 9:16, and we just use that to create a proportion.0844

So, 54:96 is going to be the same as 9:16.0855

And the last example: Use the given area to find AB.0868

So, this is what we are looking for, here: the area is given here; the area is given here.0874

This is also given; the corresponding side is given.0881

Let's label this as a and this as b; a:b would be the scale factor between the two figures.0885

We don't know a, but we know b; b is 8, so it is going to be a:8.0900

And a is what we are looking for, because that is AB.0906

Now, I know that, for the areas, it is going to be the scale factor squared; so it is a2 to b2,0911

which is a2 to...b is 8, so 82.0929

Now, that is the same thing as a2/82; so we are going to use this ratio and make it equal to these areas.0938

So, a2 is the same thing as, here, 218, over 166; so the ratio of this area to that area is a2:64.0954

And you are just going to use this proportion to solve.0971

It is going to be 166 (and I am just cross-multiplying) a2 equals 218 times 82 (is 64).0975

So, from here, you can just divide this 166; a2 =...and you can just use your calculator...218 times 64...0991

divide that number by 166, and I get 84.05.1011

And then, since we are solving for a, we need to take the square root of that;1023

so on your calculator, you can just take the square root of it; and I get 9.17.1029

So, this right here is going to be 9.17 centimeters.1039

Again, all I did was to label this a and b; the scale factor is a:8; to find the scale factor of the areas,1048

you are going to do a2 to b2, which is equal to 210966.1058

And then, solve it for the a; that is what we labeled as our AB, and that is centimeters.1070

Let me just rename this, since it is asking for AB; I'll say AB is 9.17 centimeters.1081

That is it for this lesson; thank you for watching Educator.com.1094

Welcome back to Educator.com.0000

For the next lesson, we are going to go over geometric probability.0001

The first thing that we are going to go over is the Length Probability Postulate.0008

It is when we are using segments for probability.0013

If a point on segment AB is chosen at random, and point C is between A and B, then the probability that the point is on AC0018

is going to be (and this is a ratio) segment AC, over AB.0033

Now, if you remember probability, probability measures the part over the whole.0041

You can also think of the top number as the desired outcome, over the total outcome, the total possible number of all of the different types of possible outcomes.0052

So then, it is a desired outcome, what you are looking for, over the total--over the whole thing.0077

So, it is just part over whole; this is the most basic way you can remember probability, part over whole.0083

Here, the same thing applies to the Length Probability Postulate; you are looking at the part.0091

You are looking at what the desired outcome is, which is the point being on AC, over the whole thing; that is AB--that is the whole thing.0099

It is AC to AB--always part over whole.0110

So, let's say that this right here is 5; CB is 5; the probability of a point landing on AC...what is AC?0116

That is the desired outcome; that is the top number, which is 5, over the whole thing (it is not 5; it is not the other half);0128

it is the whole thing, which is AB, and that is 10; so when I simplify this, this becomes 1/2.0139

The probability of landing on AC is 1/2.0148

And for the Area Probability Postulate, when you are talking about the probability of something to do with area, you are looking at space.0156

So, you are looking to see, for example, maybe a dart hitting the dartboard; that is area, because it is space that you are looking at.0166

If you look to see, maybe, a spinner (we are going to have both of those for our examples) landing on a certain space, that is area.0178

So, that has to do with this probability; and this postulate says that if a point in region A0191

(this rectangle is region A) is chosen at random, then the probability that the point is in region B0199

(which is inside region A) is going to be the area of region B, over...remember: the whole thing is the area of region A.0207

Region A is the whole thing; the area of the whole thing is the total,0231

and the top one will be the area of the desired outcome, or the part that we were just looking at; and that is region B.0238

The area of a sector of a circle: now, a sector is this little piece right here.0254

This is the center; the sector is the area of this piece, so it is bounded by the central angle0264

(this is a central angle right here; this angle is a central angle, so if I need the θ, that is the central angle) and its intercepted arc.0273

This is the intercepted arc; so those are the boundaries, this angle and that.0283

This whole thing is called a sector; now, I like to refer to the sector as a pizza slice.0289

Think of this whole thing as a pizza; this is a slice of pizza; so a sector is a slice of pizza.0300

We are finding the area of that slice; to find the area of this, it would be this formula:0309

the central angle (which is this angle right here, this central angle) over 360...0318

now, why is it over 360?--because going all the way around a full circle, including that, is 360; so it is like the part over the whole,0326

the central angle over the whole thing, which is 360...times the area of the circle.0337

Now, another way (an easier way, I think) would be (to figure out how to find the area of this):0346

instead of looking at this formula, I like to use proportions.0355

So, what we can do to find the area of this pizza slice right here: remember: a proportion is a ratio equaling another ratio;0359

so, we are going to look at the probability (probabilities are ratios, something to something, which is part to whole)0373

of the measures to the areas, because we are looking both: we are looking at measure, and we are looking at area.0387

So, for the measures, the part over the whole for the angle measures is going to be the central angle, over the whole thing, which is 360.0403

And the probability of the area...isn't that part over whole, also?...so the part will be the area of the sector.0422

That is the area of the sector; so let's just call that A for area of the sector...over the whole thing, which is0429

(the area of this whole circle is going to be) πr2.0440

Again, the ratio (or probability) of the part to the whole is the angle measure to the whole thing.0445

And the ratio of the areas is going to be the area of the sector, over the area of the circle, because this is part to whole.0459

We are going to make them equal to each other; that is our proportion; and you are just basically going to solve.0467

Let's say that this angle measure is 40, and the radius, r, is 6.0475

If this is 40, that is the part; that is 40 degrees, over...what is the whole thing?...360 degrees, is equal to the area of the sector;0497

that is what we are looking for; that is the area of the sector, over the area of the whole thing; that is the circle, so it is πr2.0512

So, that is π(6)2; and if you are going to solve this out, remember how you solve proportions.0524

You do cross-multiplying; so then, the area of the sector, times 360, times a, equals 40 degrees times π(6)2.0531

And when you solve this out, you divide this by 360, because we are solving for the a.0558

Now, if you look at this, this is exactly the same thing as this right here: the central angle,0568

the angle of that right there, divided by the 360, the whole circle, times the area of the circle--that is πr2.0579

It is the same exact thing; if you want to just use this, that is fine--it is the same exact thing.0591

But this way, you just know that you are looking at the part, the angle measure, over the whole, the circle's angle measure.0597

That is equal to the area of the sector (that is the part), over the area of the whole circle.0608

It is part over whole, for angle measures, equals part over whole, for the areas.0612

And that is just a way for you to be able to solve this out without having to memorize this formula.0620

And then, we solve this out; and you can just do that on your calculator; I have a calculator here on my screen.0627

I get that my area is 12.57; and again, that is the area of this sector, the pizza slice; and that is units of area, squared.0642

That is the area of a sector.0663

Let's go ahead and do some more examples: What is the probability that a point is on XY?0668

Again, for probability, we are looking at part to whole; so the desired outcome, the part that we are looking for, is XY;0675

that is going to be my numerator--that is the top part of my ratio--so XY is from 0 to 2; that is 2 units.0685

It is XY; again, we are looking at XY over the whole thing, which is XZ, so that is 2 over...the whole thing, from X to Z, is 10.0698

If I simplify this, this becomes 1/5, because 2 goes into both; it is a factor of both 2 and 10.0715

I can divide this by 2 and divide that by 2, and I get 1/5; that is the probability that a point will land on XY.0723

Find the probability of the spinner landing on orange, this space right here; here is that spinner.0736

This angle measure is 72, and the radius is 4; so, if this is 4, then we know that any segment from the center to the circle is going to be 4.0749

OK, so then, I want to use that proportion: the angle measure, over the measure of the circle, the total angle measure,0767

is equal to the area of the sector, over the whole thing (is going to be the area of the circle; and that just means "circle").0794

This is my proportion: the angle measure...any time I am dealing with the part (since it is always part over whole),0810

it is always going to be about the sector, this piece right here, the orange; and then, any time I am talking about the whole,0821

it is going to be the whole circle...(that is that) is 72 degrees, over the whole thing (is 360), is equal to0828

the area of the sector (and again, that is what I am looking for, so I can just say A for area of the sector),0842

over the area of the circle (that is the whole thing); and that is πr2.0850

My radius is 4, squared; so then, I can go ahead and cross-multiply.0859

360A = 72(π)(42); then, to solve for A, divide the 360; divide this whole thing by 360.0870

And then, from there, you can just use your calculator: 72 times π times the 42...and then divide 360; you get 10.05.0891

And we have inches here for units, so it is inches squared; that is the area of this orange.0919

Now, to find the probability...we found the area of this orange; and be careful, because,0926

if they ask you for the area of this base right here, then that would be our answer; but they are asking for a probability0933

of landing on orange; and any time you are looking at probability, you are always looking at part over whole.0939

And again, since we are talking about area, it is the area of the orange, over the area of the whole thing.0947

I found the area of a sector; now, to find the area of the whole thing, the area of the circle is πr2.0958

And all you have to do is...we know that r is 4, so 16 times π is 50.27 inches squared; that is the area of the circle.0970

And then, the probability is going to just be (I'll write it on this side) 10.05/50.27.0990

You can change this to a decimal, so you can go ahead and divide this; or maybe you can just leave it like that,1007

depending on how your teacher wants you to write the answer.1015

You can definitely have probability as a decimal; you can just go ahead and take this and divide it by this number; and that would be your answer.1020

This is the probability, part over whole, the area of the orange over the area of the circle.1028

The circle is circumscribed about a square; if a dart is thrown at the circle, what is the probability that it lands in the circle, but outside the square?1040

We want to know what the probability is of landing in the gray area: it said "in the circle, but outside the square"; that is all the gray area.1055

That is the probability: they are not asking for the area of that part; they are asking for the probability of landing on that part.1066

So then, we have to make sure that we are going to do the part over the whole.1073

First, I have to find the area of that gray area, because that is my desired outcome; that is my part.1082

The desired outcome is the area of the gray, over the whole thing, which would be the area of the circle, because that is the whole thing.1087

So then, my part is going to be, again, area of gray over the area of the circle.1097

To find the area of the gray region, we have to first find the area of the circle and subtract the area of the square.1117

The area of the gray is going to be the circle, minus the square.1132

The area of the circle is πr2, minus...the area of this is going to be side squared.1154

We know that the radius is 10, because, from the center of the circle to the point on the circle, it is π(10)2;1168

minus...do we know what the side is?...we actually don't, because this is from the center to the vertex of this square.1180

So, let me make a right triangle: I know that this angle right here (let me just draw the triangle out again--that doesn't look good;1191

this is more accurate)...this is that triangle here: this is 10, and I want to know either this or this.1208

Let's say that we are going to call that x.1219

Now, this is a right angle; we know that this is a 45-degree angle, because it is half the square; in squares, everything is regular.1223

So, to find the other sides of a 45-45-90 degree triangle, since we know that it is a special right triangle, we are going to use that shortcut.1236

If this is n, then this is n, and this is n√2; and in this case, I should label this n, because that is the side opposite the 45, which is n.1250

The side opposite this 45 is n, and then the side opposite the 90 is 10.1268

Here is the shortcut; I am given the 90-degree side, which is this right here, so I am going to make those equal to each other,1273

because this is n, and this is n√2, which is 10; so n√2 = 10.1281

Divide the √2 to both sides: I am going to solve for n; n = 10/√2...what do I do here?1292

Well, this square root is in the denominator, so I have to rationalize it; when I do that, this becomes 10√2/2; simplify this out; this becomes 5√2.1303

So again, what did I do? I took this...because I have this right here, the hypotenuse of this right triangle, I want to find this side right here.1324

I am going to use special right triangles, since this is a 45-45-90 degree triangle: n, n...the side opposite the 90 is n√2.1332

That is the side that I am given, so I am going to make that equal to n√2: n√2 = 10.1343

Solve for n by dividing the √2; let me rationalize the denominator, because I can't have a radical in the denominator.1350

So then, this becomes 10√2, over...√2 times √2 is just 2; simplify that out, and I get 5√2.1360

That means that n is 5√2; this side is 5√2; this side is 5√2.1370

Well, if this is 5√2, then what is this whole thing? We labeled that as s.1382

So, if this is 5√2, then this is 5√2; so you basically have to just multiply it by 2, because this is half of this whole side.1389

My side is 5√2 times 2, which is 10√2.1400

So, to find the area of the square, I am going to do 10√2 times 10√2, base times height (or side squared): 10√2, squared.1412

And then, I am going to use my calculator: this is 3.14 times 100, which is 314, minus 10√2 times 10√2;1427

that is 100 times 2; that is 200 (if you want, you can just double-check on your calculator).1445

This right here is (I'll just show you really quickly) 10√2 times 10√2.1452

10 times 10 is 100; √2 times √2 is times 2, so it is 100 times 2, which is 200.1464

Then, this is going to be 114; so the area of the gray is 114, because I took the area of the circle,1478

which was πr2, 314, and then I found the area of the square, which is 200, 10√2 times 10√2.1491

And then, I got 114; now, that is just the area of the gray; we are looking for the probability that it lands in the gray.1510

That is the area of the gray, over the area of the whole thing, which is the circle.1523

We take 114 over the area of the circle (where is the area of the circle?), which is 314; and that is the probability.1528

Now, we know that both of these numbers are even, so I can simplify it.1546

So then, if you were to cut this in half, this is going to be 57; if you cut this in half, this is going to be 157.1552

And that would be the probability, 57/157.1564

So again, the probability is going to be the area of the gray over the area of the whole thing, which is the circle.1573

The fourth example: we have a hexagon, and I am just going to go ahead and write that this is a regular hexagon with side length of 4 centimeters.1586

It is inscribed in a circle; what is the probability of a random point being in the hexagon?1599

"Inscribed": now, I don't have a diagram to show you, so I am going to have to draw it out.1606

"Inscribed" means that it is inside, so the hexagon is inside the circle; but all of the vertices of the hexagon are going to be on the circle.1611

They have to be intersecting; so let me first draw a circle, and the regular hexagon ("hexagon" means 6 sides).1627

I am going to try to draw this as regular as I possibly can, something like that, so it will look like it is inscribed...something like that.1641

What else do we have? Side length is 4 centimeters; what is the probability of a random point being in the hexagon?1666

OK, so then, again, we are giving a probability, so it is part over whole.1679

What is the part? The part will be the hexagon; inside the hexagon is the desired outcome--that is what we are looking for.1686

So, it is going to be the area of the hexagon, over...the whole thing is going to be the area of the circle.1693

Let's see, now: to find the area of this hexagon...remember: to find the area of a regular polygon,1711

if we were to take this hexagon and then break this up into triangles, we have 1, 2, 3, 4, 5, 6 triangles.1727

Each triangle is going to be 1/2 base times height; and then, we have 6 of them...times 6.1743

Now, if we take the base (the base is right here), and we multiply this base with this 6, isn't that the same thing as the perimeter?1756

The base, with the 6, is going to be the perimeter; the height, this right here, we call the apothem.1766

I am going to draw arrows to show that the base and the 6 together became the perimeter, and the height became the apothem.1785

1/2 just stayed as 1/2; this is the formula for the area of a regular polygon: it is 1/2 times the perimeter of the polygon, times the apothem.1792

The apothem is from the center, the segment going, not to a vertex, but to the center of the side; so it is perpendicular.1808

Now, we don't know what the apothem is, so I am going to have to look for it.1824

Now, remember: you always want to use right triangles, if you possibly can; we can, because the apothem is perpendicular to the side.1833

So, if I just maybe draw it bigger to show: this is the apothem.1843

If the whole side measures 4 (see how this is 4), then this half is going to be 2.1853

And then, I want to look for this angle measure, because I don't have this side.1863

If I have this side, then I can use the Pythagorean theorem, because then I have a2 + b2 = c2.1868

But I don't, so instead, I want to see: this is a circle; the whole thing, all the way around, is 360 degrees.1876

If I break this up into parts, this is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.1886

I basically just look to see how much this triangle is from the whole 360; if the whole thing measures 360,1901

remember how we said that this is actually one of the 6 triangles; so 360 divided by 6 is 60.1913

Then, since this whole thing measures 60 degrees, what is this half right here?1927

This half is 30 degrees, so this is 30 and this is 60; and again, all I did was just...1932

I know that the whole thing, the full circle, measures 360; since I know that this right here is 60,1942

because I have 6 triangles, so it is like 6 triangles sharing 360 degrees;1950

then this angle right here is 60 degrees, which means that this half right here is 30.1956

A 30-60-90 triangle is a special right triangle; if this is n, the side opposite the 60 is n√3; the side opposite the 90 is 2n.1964

This is the special right triangle; what do I have--what am I given?1981

The side opposite the 30 is 2; so I am going to make those two equal to each other: n = 2.1988

That means that the side opposite the 60 is going to be 2√3; the side opposite the 90 is going to be 2 times n, which is 4.1999

That means that a, which is the side opposite my 60, is 2√3; isn't that my apothem, my a?2012

So, I know that that took a while; but it is just going over the area of a regular polygon.2020

It is 1/2 perimeter (which is 6 times 4, because there are 6 sides, and each side is 4...so perimeter is 24)...my apothem is 2√3.2033

And then, I can just cut this in half; so 24/2 is 12, times 2 is 24, √3.2053

To make that into a decimal, 24 times √3...we get 41.57 units squared...oh, we have centimeters, so this is centimeters squared.2065

And again, this is the area of the hexagon; so this is the hexagon, and then I want to find the area of the circle.2094

Now, if it were just the area of the hexagon that we were looking for, then this would be the answer.2110

But again, we are looking at probability: what is the probability of a random point being in the hexagon?2113

It is the area of the hexagon, over the area of the full circle.2124

So then, I look at it: here is the hexagon; here is the circle; the area is πr2.2131

π...do I know r?...r would be this length right here; this is the radius, because this is the center of the circle; this is a point on the circle.2144

From here to here...if this triangle is this triangle...what do we find about this side?2159

That side is opposite the 90, and that is 4; so this is 4 squared; this is 16π; 16 times π is 50.27 centimeters squared.2168

So then, hexagon over circle is 41.57, over 50.27; so let me just do that on a calculator: 41.57/50.27...and I get (so the probability is) 0.83.2199

Now, one thing to mention here: when you have a decimal, when you change your probability fraction into a decimal,2239

you have to make sure that it is less than 1, because, if you are looking at a part over the whole,2247

it is going to be a proper fraction; the part is going to be smaller that the whole.2257

If the whole is everything--it is the whole thing--well, then, the part can only be a fraction of it.2262

So, the only time you can get anything greater than this...2269

I'm sorry: the biggest number you can get for probability, when you change it to a decimal, is 1, because,2276

when you look at the fraction, it can just be the whole thing over the whole thing.2284

And when you have whole over whole, well, that is just going to equal 1, because it is the same number over itself.2289

So, make sure that your probability decimal is not greater than 1, unless you are talking about the whole thing.2296

Then, it is going to be 100%, all of it, which is 1; but otherwise, if the part is smaller than the whole, then your decimal has to be less than 1.2304

That is it for this lesson; thank you for watching Educator.com.2318

Welcome back to Educator.com.0000

For the next lesson, we are going to go over three-dimensional figures.0002

Another word for three-dimensional figures is solids; now, some specific types of solids are polyhedrons.0007

Whenever you have a three-dimensional solid with all flat surfaces, meaning that all of the sides of that solid are flat, then it is a polyhedron.0018

These are examples of polyhedrons: notice how all of the sides are flat--no round surfaces; no circles,0031

because if you have circles as a side, then the other sides that are intersecting with that are not going to be flat.0042

We are going to go into those types of solids more later.0050

But with these, notice that all of the sides are flat, and it has to enclose a single region of space,0054

meaning that it can't be open; it can't be that one side is missing--it has to be closed.0063

The three parts of polyhedrons: the first one are faces; now, the faces are the sides.0069

All of this right here would be a face, and the same thing with all of this, and this is a face...0080

Each one of those is a face--any time you have a side.0095

This one here has four faces: you have 1, 2, 3, and then the one in the back that you can't see; so each side is a face--these are sides.0099

Edges are the segments that are the intersections of the faces.0116

It is where the faces intersect, so these are all edges here; they are segments.0130

And the vertices are all of these; they are the corners, the endpoints of the edges; so these are like corners, the corners of these polyhedrons.0146

Going over solids: the pyramid is the first one: we know that it is a polyhedron, because all of the sides are flat surfaces.0174

So, all of the faces and one base...the base is the bottom part of this pyramid, so if I draw that part,0186

it is considered the base, the bottom part; and each of these sides here is a lateral face.0205

And all of those faces meet at a point, and that is the vertex.0217

We are going to go over each of these more specifically later on.0222

The next is a cylinder; a cylinder is like a can, like a can of soup; it is a cylinder.0226

The bases would be the circles on top and the bottom, so that would also be the base.0236

A cone is one circular base, and a vertex is right here; think of an ice-cream cone.0252

A sphere is like a ball; and this is not called a circle, because circles are two-dimensional; this is a sphere--it is three-dimensional.0265

A basketball would be a sphere; it is a set of points in space that are a given distance from a given point.0279

This point right here is the center of the sphere; and no matter which way you go, no matter which direction, it is always going to be the same distance.0286

The next type of solids is a prism; a prism is when you have two bases that are opposite and congruent, and they are parallel.0300

This is called a rectangular prism; in this case, I can name any two sides bases,0315

because as long as the two opposite sides are parallel and congruent, they would be the bases of a prism.0325

Now, another example of a prism would be like this; now, if you notice, this is not rectangular.0337

So, this would be the base, and this would be the base opposite; these two bases are parallel and congruent, so this is a prism.0366

Now, the base itself is not a rectangle...this would be called a rectangular prism, because the base is rectangular in shape;0382

and for this one, this is a special type of prism because, if I choose this as the base,0398

the side opposite is also going to be parallel and congruent, so these two can be named the base.0409

Or if I want, I can make the top and the bottom the base, or I can make the front and the back the base,0423

because, as long as, in your prism, you have two opposite faces that are parallel and congruent, they can be the bases of your prism.0435

And that would be the type of prism.0446

Two of the faces are bases; the rest of the faces are lateral faces.0451

So in this one, this top and bottom are your bases; the rest of the sides (that means this one right here,0458

this one right here, the one in the back, and then the three other sides) would be your lateral faces.0466

So, bases and lateral faces make up all of the sides, or faces, of the prism.0474

A regular prism is when the bases of your prism...so then, for this one, if I am going to name0483

the top and the bottom the base, they have to be regular.0493

So, as long as the bases are regular (and regular, again, means equilateral and equiangular), so if I am going to call0497

the top and the bottom the bases, if all of these sides are congruent,0516

and all of the angles are congruent, then this would be considered a regular prism.0521

For this one, if this pentagon (because that is the shape of the base) was regular, then this would be considered a regular prism.0531

And if this is regular, then we know that the bottom base is regular, because they are the same.0545

If the base is a regular polygon, then the whole thing, the prism, is a regular prism.0552

And a cube is when each side of a rectangular prism is a square; so that means the front is a square;0559

the top side is a square; the side is a square...if all of these are squares, then this is a cube.0572

Platonic solids: there are five types of platonic solids, and they are all regular polyhedra.0590

"Polyhedra": if we have one, it is a polyhedron; but when we have more than one, it can be either polyhedrons, with an s to make it plural, or polyhedra.0599

So here are the five types: this one right here has 4 sides, the front, this side...there is a back side, and there is the bottom base.0616

So, it will look like this: 1, 2...and then this is the back side that you can't see...0630

So, the one with four sides (and again, this is all regular, meaning that the sides are all regular) is a tetrahedron.0646

The one with 6 (this is a cube, the one with 6 sides) is a hexahedron.0667

This one has 8 regular sides, so this is an octahedron, just like an octagon.0679

This one has 12 sides, so this is a dodecahedron.0693

This one has 20 sides, and that is an icosahedron; those are the five types of regular polyhedra.0707

OK, slices and cross-sections: whenever you have a solid, and you have a plane that intersects the solid, then you have a slice.0728

"Slice" sounds like you are cutting through, so it is intersecting; so here is a cone, and here is a plane intersecting the cone.0743

See how it is cutting through it...so that is a slice.0752

Then, where they intersect, that is the shape that is formed.0759

Now, if you have the plane intersecting the solid so that the plane is parallel to the base0768

(in this case, the plane is parallel to the base; they are both horizontal, so the base is flat like this,0778

and then the plane is flat or horizontal, cutting through somewhere in the middle), then that would be a cross-section.0786

So, when it is parallel, when the parallelogram is slicing so that it is parallel to the base, then it is a cross-section.0798

So in this case, this would be a cross-section.0808

And the figure, or the shape, that is formed by the intersection would be a circle.0811

Now, you can see here that it is like this; that is where they are intersecting.0817

But that is because it is kind of sideways; if you were to draw it from a top view,0823

meaning if you had a counter right here and you are looking down this way, then you would see a perfect, round circle.0827

That is not perfect, but it is a circle; so that is the shape that resulted from the slice, from the intersection of the plane and the cone.0835

So then, again, when the plane intersects the solid, it is slicing it, and the shape that resulted from that is a circle.0848

Moving on to our examples: name the edges, bases, and vertices of the polyhedron.0865

Remember: edges are all of the sides, all of the segments where all of the faces intersect.0873

So, for the edges, I have to just name all of these segments; so I have AB; I have AE; let me just do that,0884

so that I know that I wrote it down; AE, and then there is that one, BE, this one, AF, BC, ED0905

(it doesn't matter if you do ED or DE), CD, CF, and FD; so then, I have 1, 2, 3, 4, 5, 6, 7, 8, 9--those are all of my edges.0928

And then, bases would be triangle ABE and then the triangle CFD.0959

Then, I am going to name this side here (that is ABCF), then this one here, which is AFDE, and (which one?) this one right here on this side; that is BEDC.0985

Make sure that, when you label it, you label it in order: BEDC--it can't be BDCE--it can't go out of order (but you can do BCDE).1014

And then, we have vertices: vertices would be just these corners, so it would be vertex A, vertex E, vertex B, vertex F, vertex C, and D.1029

Those are the edges, faces, and vertices.1053

The next example: Determine if the figure is a polyhedron, and explain why.1058

This is a cone; it has one base, which is a circle, and we have a vertex.1064

Now, this is a solid, because it is three-dimensional; but it is not a polyhedron, so this would be "no," because polyhedrons have only flat surfaces.1074

Now, this base is flat; it is a flat circle; but how about the cone--isn't it circular? It is not flat.1094

There is no side to this, so it is "no," because it is not a flat surface.1105

So, the solid, because there are no faces, does not have flat surfaces as the sides.1116

Even though, again, the bottom one, the base, is flat, the side here is not flat; it is going circularly.1140

OK, the next example: Describe the slice resulting from the cut.1154

As you can see here, this cone is being cut, or sliced, sideways, like this; so the parallelogram is cutting into the side.1159

Now, what is the shape that has resulted from this cut--what is that cut going to look like, where they intersect?1172

Well, if you have a cone here, it is going to go like this; it is cutting right there--that is a part of it;1184

and then, what about the other side?--it is probably going to go like that, something like this.1197

So, this shape would be a parabola; it is going to be like a U--something like that; that is the shape that is from the slice,1207

from the intersection of the plane and the cone.1224

Just another example: if you have, let's say, a cylinder, and let's say you have a plane that is intersecting,1232

or slicing, that cylinder; well, what is the shape that resulted from the slice?1246

Again, if you look at it from a top view, it is going to be a circle.1261

Think of a can, and you cut in half down the middle; you get a knife, and you just cut it through the middle; aren't you going to get a perfect circle?1272

So, that is another example of a slice.1280

Describe the shape of the intersection: A cube is intersected by a plane that is parallel to the base.1288

A cube is being intersected so that it is parallel to the base; so then, my plane will be something like that.1298

It is being intersected right here; and it is parallel to the base, so that would be a perfect square.1328

A sphere is intersected by a plane at the center of the sphere.1349

I have a sphere; I don't know how you would show a sphere; and it is being intersected right here by a plane.1355

As a result, it is going to look like this; and it has to cut through the center, so what is the result from this slice?1381

It would be a perfect circle, like that.1395

Again, if you look at it from your screen, you are going to see that it is going to look like this; but you have to look at it from a top view.1401

And it is going to be a perfect circle.1412

And that is it; thank you for watching Educator.com.1416

Welcome back to Educator.com.0000

For the next lesson, we are going to go over the surface area of prisms and cylinders.0002

First, let's talk about prisms: prisms, we know, are solids with two bases that are both parallel and congruent.0008

They are congruent polygons that lie in congruent planes; they have to be parallel and congruent.0025

If you look at this prism here, this top and this bottom right here would be the bases; they are both parallel and congruent.0038

Now, the other sides, the remaining faces, are lateral faces; they are all of the faces that are not the bases.0054

And each of them, each of those lateral faces, would be a parallelogram.0064

So, if you notice, this is a parallelogram, parallelogram, parallelogram; all of them will be parallelograms.0070

Lateral edges are those line segments where the faces intersect; so this face and this face intersect here--those are lateral edges.0080

And the lateral edges would be only from all of the lateral sides, where the lateral sides intersect--not these.0096

These are considered edges, too, but they are not lateral edges, because these are edges from the base.0106

It would be all of these here; those are lateral edges.0113

And the altitude measures the height of the prism; so altitude, we know, has to be perpendicular to the segments.0120

This perpendicular segment is the altitude; it measures the height, because height, we know, has to be perpendicular.0131

The two types of prisms: it is going to be either right or oblique.0146

A right prism is when these lateral edges are altitudes; so the measure of the lateral edge, the length, is the height of the prism.0151

So, this right here measures the height; then it is considered a right prism, because it is standing up (right).0167

An oblique prism is one that is tilted; it is slanted to the side.0177

So, in this case, these lateral edges are not perpendicular; they are not considered altitudes, so they are oblique; that is a prism that is not right.0181

So then, if you have to find the height of this, then you would have to find the height that is perpendicular to the base.0195

Classifying prisms: classify a prism by the shape of its bases.0209

Depending on the bases, there are different names for these prisms.0216

These are all prisms; all four of these are just a few types of prisms.0220

The first one: if we label the top and the bottom as bases...0227

now, for this rectangular prism, it is a special type of prism, because we can actually name any pair of opposite sides as its bases,0235

because we know that bases just have to be congruent and parallel;0245

and for this, each pair of opposite sides is congruent and parallel.0251

So, for a rectangular prism, it doesn't matter which two opposite sides you label as bases.0256

But if it is standing this way, then just to make it easier, you can name the top and the bottom as bases.0263

Then, this would be a right (because it is standing upright; all of the lateral edges are perpendicular) rectangular prism.0270

This one here...now, be careful here, because in this case, the bottom is not considered the base.0297

This triangle right here, with this front side and this back side, would be considered the bases.0310

The other sides, the bottom, the right side, and the left side, are all lateral faces.0327

So, don't always think that the top and the bottom are going to be the bases; in this case, it is the front and the back, and they are triangles.0336

The bases are triangles; that would make this...0345

and for determining whether it is right or oblique, if we were to take this solid and stand it up,0349

so that the bases were the top and the bottom sides, then this would be the height; it would be the altitude.0357

And it would be perpendicular to the bases; so this is also a right...0367

and then, the shape of the base is a triangle; so this is a triangular prism.0374

This one here, we know, is oblique, because we can just tell that it is not standing up straight; it is slanting to the side, so this is oblique.0388

And then, the bases would be this top and the bottom; and they are 1, 2, 3, 4, 5...5 sides, so that is a pentagon: pentagonal...this is "al"...prism.0404

And this last one here, we know, is right; and how many sides is the base?0435

We know that this top and the bottom is a base, again; 1, 2, 3, 4, 5, 6...so that is a hexagon, so that is a hexagonal prism.0443

OK, to find the lateral area of a prism, first make sure that you determine which sides your bases are and what your lateral faces are.0464

And once you do that, your lateral area is just the area of the lateral faces.0478

If we say that the top and the bottom are the bases, that means that we are finding the area of all of these four sides.0487

Left, front, right, and back--we are looking at the area of all the four sides, minus the bases--not including the bases--and that is lateral area.0498

The formula to find the lateral area of a right prism would be the perimeter of the base, times the height.0516

And the reason for this formula is (I am going to explain it to you): let's say that we take scissors, and we cut one of these sides.0524

Let's say you cut it there--cut that corner--and you unfold it.0545

When you unfold it (and I have a paper here to demonstrate), here is the rectangular prism.0551

We know that it is a rectangular prism, because the bases are a rectangle.0565

If you were to cut it, and you unfold, you get a rectangle; so again, this is a rectangular prism;0572

if you cut it and unfold, then you get a rectangle.0592

This is my cut that I made; you end up getting just a big rectangle.0605

So, if this is side 1, this right side; this front side is side 2; this is side 3 on the left; and the back is side 4;0615

well, it is as if I have side 1, side 2, side 3, and side 4; and actually, this cut is made to this side, the left side, or the right side of side 1.0627

So, it is as if this would be side 1; so either way, it is just a big rectangle.0651

So, each of these sides, 1, 2, 3, 4, 1, 2, 3, 4...then, when you fold it back up, it will be like this, with side 2 folding this way, this way, and this way.0659

It is as if you are taking this, and you are folding it back over.0671

Now, in that case, since the lateral faces all make up the big rectangle (I am going to erase this, so you don't get confused),0676

let's say that this right here has a measure of 2, and let's say that each of these is the same;0702

well, then I know that this is 2; this is 2; 2; and 2; and the height, the altitude, is, let's say, 10.0711

Then, I know that this right here would be 10.0724

The perimeter here: to find the area of this, it would be all of this length right here, times this.0730

Here, this is 2, 4, 6, 8; this has a measure of 8, and this is 10; so the area of all of these lateral faces would be 80.0741

So then, that is how this formula came about, because, if you were to cut it, well, then the area of all of the lateral sides0754

would be the perimeter of the base (because you are doing this, this, this, and this) times the height,0766

because if you were to unfold it, this, that, and all of these sides would come up to 8.0772

And isn't that the perimeter of the base?0779

So, that is why it is the perimeter of the base, which is this, times the height; and that is 10.0787

So then, the lateral area of this would be 80 units squared, because it is still area; you are finding all of the space.0803

That is 80 units squared, and that would be only the lateral area.0814

Now, next, let's go over surface area: the surface area is the area of all of the sides, so it would be the lateral area, plus the area of the two bases.0824

It is the lateral area, plus 2 times the area of the base.0845

This top right here is the base, and then this right here is also a base.0851

Now, if you want, you could find the area of each one of these: 1, 2...all of the sides, all of the bases, and then just add them all up.0858

That would give you the surface area, because it is the sum of the areas of its outer surfaces--all of the sides.0869

But since we know that the lateral area would be the perimeter--so again, if you were to cut it,0877

let's say, right down here, and unfold the lateral area, then it would just be the perimeter,0885

because it would be this side, this side, this side...all of those sides, which is the perimeter, times the height.0895

It is going to just give you one big rectangle; the perimeter of the base is going to be the length; so this is the perimeter of the base,0911

and this is the height; so that, plus...and then the area of the base, times 2, because you have 2 bases.0933

And then, that would give you the area of all of the sides together.0951

That is surface area: lateral area is just the area of all of the lateral sides, and then the surface area would be0958

the area of all of the outer sides, including the bases; that is lateral area, plus the area of the bases.0967

Next, we have cylinder: a cylinder has two bases that are parallel and congruent circles.0979

So, it is like the prism, except that the bases are circles instead.0986

Here is the base here, and here.0990

The axis is the segment whose endpoints are the centers of the circles.0994

So, it has to go from the center of one base to the center of the other base; that is the axis.1003

Now, it could be different from altitude; now, in this case, in a right cylinder, the axis is the same as the altitude, because the altitude measures the height.1013

If it is standing up straight, then it doesn't matter where the endpoints are--center to center or from end to end.1024

As long as the endpoints are on the two circles, that is the altitude.1037

For a right cylinder, altitude is the same thing as the axis; in an oblique cylinder, that is not the case.1042

The altitude is right here; that measures the height; this is the altitude.1056

But the axis, remember, has to go from the center to the center of the two circles; so this right here is the axis.1064

It is not the same in an oblique cylinder.1077

Now, the lateral area of a cylinder is the same concept as the lateral area of a prism; lateral area is the same.1084

To find it within the cylinder, again, it is just the area of everything but the bases.1094

So, we know that this is the base; we know that this is the base; so it would be the area of just the outer part.1103

Now, think of a can, like a soup can; and you tear off the label--the label goes around the can.1113

It is like finding the area of that label; that would be like lateral area.1121

Again, if you make a cut, like this paper here--if you have a cut--there is your cylinder without any bases,1129

and you cut it, and then you open it up, you are going to get a rectangle.1147

To find the area of the rectangle (the lateral area just means that you are finding the area of a rectangle),1163

you need base, and you need height; the base would be (if I turn it back into a cylinder)...1169

isn't the base...let me actually call it the width--the width and the length--so that you don't get it confused with these bases.1182

The width of the rectangle is the same as the measure of this other circle.1198

What is that called? That is called circumference: to measure this all right here, that is the circumference.1206

The width of this rectangle is 2πr, because it is the circumference--just 2πr.1218

And then, the length, we know, is h; it is the height of that.1233

That is how we get this formula here: 2πr times the height--that is the lateral area.1241

And for surface area of a cylinder, the same thing works: lateral area, plus the area of the two bases.1255

Now, this is the easiest way to remember it, because the lateral area is always just going to be a rectangle.1268

And then, we just find the area of the circle, and the area of the other circle--1277

or find one of them, and then, since they are going to be the same, just multiply it by 2,1285

and then add them together: so it is this area, times 2, plus that; all of these together is going to equal the surface area of the cylinder.1290

It is going to be πr2, so it is 2 times πr2.1303

Let's do some examples: Find the lateral area and surface area of the prism.1312

Now, the first thing to do is to figure out what the bases are.1319

Whenever you have a prism, the easiest way to point out the bases is to look for any shapes, any sides or bases, that are not rectangular.1329

Automatically, we know that these two triangles will be the bases.1342

Now, if you have a face that is not rectangular, but there is only one of them, then it can't be a prism; it is going to be something else.1351

It will probably be a pyramid or...I don't know; it is not going to be a prism.1360

If it has two opposite sides that are congruent and that are not rectangular, then those two sides would be the bases of the prism.1369

Those are my bases; and then, the lateral area--imagine the cut; you are going to unfold; it is going to be one rectangle.1384

And that is three sides, three lateral faces, that make up the rectangle, this lateral area; the three sides go like that.1396

One side is 6--that is one side; the other side is 6, because this side and this side are the same; and then, the other side will be 5.1412

So, if this is where the cut is made, then it is as if this is where the cut is made; so 6 + 6 is 12, plus 5 is 17;1434

so this whole thing right here is 17, and then the height is 8.1461

We are going to do 17 times 8 to find the lateral area: on your calculator, do 17 times 8, which is going to be 136 units squared.1470

That is lateral area; then, surface area--all I have to do for surface area, since I have my lateral area,1496

is to find the area of the base, multiply it by 2 (since I have 2 of them), and then add it to this lateral area.1501

Find the area of the triangle: now, to find the area of this triangle, it is going to be 1/2 base times height; this is 6; this is 6; and this is 5.1510

I need the height; now, for the height, because this is an isosceles triangle, I know that this is half of this whole thing,1525

so this will be 2.5, half of 5; then, to find the height, to find h (this is 2.5, and this is 6), I can just use the Pythagorean theorem.1539

So, h2 + 2.52 = 62; 2.5 squared...h squared is 29.75;1554

take the square root of that, and my height is 5.45.1580

1/2...my base is 5, and then, for this base, make sure that it is the whole thing; it is not just this half.1596

We only use that half just to look for the height; we can use the Pythagorean theorem for the height and make this a right triangle.1611

But when it comes to the actual triangle, we are finding the area of the whole thing, so you have to use 5 as the base.1618

And the height is 5.45; I am just using my calculator that I have here on my screen; I get that my area of this triangle is 13.64.1624

My surface area is going to be 136 + 2 times the area of the triangle, and I got 13.64.1652

Multiply it by 2, and add it to 136; and I get 163.27 units squared.1668

That is the lateral area and surface area of this prism here.1686

The next example: Find the lateral area of the prism.1697

Now, this doesn't look like a prism; it looks kind of odd-shaped.1700

But remember: as long as you have two sides that are opposite and congruent, it doesn't matter what shape it is; those are the bases of the prism.1705

In this case, for this solid, we have the front and the back as the bases; this whole thing right here is considered the base.1719

So, that way, each of the lateral faces is rectangular; they are all rectangles.1732

Again, if you were to take this solid and stand it up, so that the bases would be the top and the bottom,1743

and then make a cut like this, it is going to be the perimeter...if you unfold it, it is one long rectangle;1753

and then, this right here, the length of the lateral area, is going to be the perimeter of the base.1769

So, it is going to be like this--all of this is going to make up this whole thing right here.1778

There is our cut; then this is like having a 1; and this is 5; that is this and then this, and then that would be the 4; and so on, all the way through.1785

I am going to do perimeter as 1 + 5 + the 4 + 8 +...and then, what is this side here?1809

This side would be the 4 plus the 1; this is 5, plus...and then, this whole thing right here is the 8, plus the 5; that would make this...and that is 13.1825

The perimeter I get: this is 6, plus 4 is 10, plus 8 is 18, plus 5 is 23, plus 13 is 36.1843

So, this is 6; this is 10; this is 18; this is 23; and then, together, they are 36.1859

And so, after that, we need to find the height; this whole thing right here is 36, and then what is this height right here?1869

That is 2, because all of these have to be the same; so my lateral area is going to be 36, my length, times the width, and that is 2;1881

so, that will be 72 units--whether it is inches, feet, and so on--squared.1900

And then, let's find the surface area of that same figure.1916

The lateral area was 72 units squared; then I want to find the area of my base, because, remember:1925

surface area would be the sum of all of the sides, so it will be lateral area, plus the area of the base, plus the area of the other base.1937

So, it is 2 bases; we have to add both to the lateral area to get surface area.1949

Here, to find the area of this base right here (just this front--this is a base), I need to break this up,1955

because there is no way that I can find the area of that, unless I break it up into 2 polygons, like that.1965

Here, this will be 8 times...what is that?..this is 4, and then this is 1, so then this would be 5.1975

So again, length times width here--this is 40 units squared, and then, for this right here, it would be 5 times 1; so that is 5 units squared.1987

Then, I add these together, and this would be 45; so the area of the base is 45 units squared.2006

But then, since I have two of them--I have a front, and I have a back--my surface area is going to be my lateral area,2020

all of that, plus two times my base; so that is 45; so 72 +...2(45) is 90...2028

and then this will be 162 units squared; this is my surface area, then.2044

The fourth example: Find the lateral area and the surface area of the cylinder.2060

Now, in the same way as our prism, if we make a cut right here, and lay it out flat, then it will just be a rectangle,2065

whereas this is the circumference, because this measures from here all the way around here, and that is the circumference.2086

So, it is 2πr, and then this is the height, which is 9.2098

2πr is 2 times π times...the radius is...4; that is the length; the width is 9.2107

Multiply that together; 2 times 4 times 9...you get 226.19...now, I am just rounding to the nearest hundredth, 2 places after the decimal.2126

That would be the lateral area; now, you can probably just leave it in terms of π, if you can.2169

2 times 4 times 9: 2 times 4 is 8, times 9 is 72; so you can probably leave it as 72π units squared.2179

But otherwise, if you have to solve it out, then you can just use your calculator: 72 times π, which is 3.14.2193

This would be the answer for the lateral area.2201

And then, to find surface area, we are going to find the area of the base, which is a circle.2205

The area of a circle is πr2; π times r, the radius, is 4 squared, which is π, or 3.14, times 16.2213

16 times π is 50.27 units squared, and that is the area of one of the circles.2237

But since I have two of them, I need to multiply this by 2; so my surface area is my lateral area, plus 2 times the area of the base.2255

And I am going to put a capital B there, to represent the area of the base:2272

this is 226 (I am going to use this number up here) and 19 hundredths, plus 2 times 50.27.2277

And then, using your calculator, solve that out; and you should get 326.72 in...don't forget...units squared; that is the surface area.2292

OK, well, that is it for this lesson; thank you for watching Educator.com.2327

Welcome back to Educator.com.0000

For the next lesson, we are going to go over surface area of pyramids and cones.0002

First, let's talk about pyramids: this is a picture of a pyramid; we have a vertex right here;0009

all of these lateral edges meet this vertex; all of the faces touch at the vertex.0021

The base is the only side, or only face, that is not going to intersect at the vertex.0037

It is going to connect all of the lateral faces together; so all of these sides here are all of the lateral faces, except the base,0046

just like all of the other solids that we went over; we have lateral faces, and then we have bases.0053

Pyramids only have one base; lateral faces always form triangles.0059

Because we only have one base, and all of the lateral faces meet at the vertex, they are all going to form triangles.0068

Lateral edges have the vertex as an endpoint; so again, one of the endpoints of each lateral edge is going to be the vertex.0080

The altitude is the segment from the vertex perpendicular to the base.0095

So, the segment from the vertex to the base so that it is perpendicular is the altitude.0099

A regular pyramid is when the base is a regular polygon; so this is going to be a square,0114

because a rectangle that is equilateral and equiangular is going to be a square.0127

So, the base is regular; the endpoints of the altitude are the vertex and the center of the base; that is the altitude--only in a regular polygon.0132

All of the lateral faces are congruent isosceles triangles, because it is regular; from the vertex,0150

it is going to be the altitude, and then we have each of these sides of the polygon, of the base, being congruent.0162

And that is going to make all of these triangles (because, remember, all of the lateral faces are triangles) isosceles triangles.0168

So, this and this are going to be the same; from this triangle right here, each one of them is going to be isosceles.0175

The slant height is the height of each lateral face; the slant height is not the same as the height of the solid.0186

Here, this height, from the vertex down to the center--we call that altitude.0199

The slant height is the height of the lateral face, the polygon; so if the lateral face is like this, this is the slant height right here.0206

Remember: it is not going to the center of the pyramid; the slant height is to there--this would be the slant height.0225

See how, when you just look at the triangle itself, the polygon, the face, it is the height, like this; it is the height of that polygon.0242

But when you look at it as a solid, the whole thing as a pyramid, it is slanted;0255

so be careful there to distinguish between the slant height and the height of the pyramid.0262

To find the lateral area of a pyramid, it is going to be 1/2 times the perimeter of the base, times the slant height--0275

meaning not the altitude, not this height; it will be the slant height, the height of the triangle.0286

Now, to try to make sense of this formula, we know that lateral area would be the area of all of the faces except the base.0294

So, for this one right here, I have four lateral faces: I have this front; I have this right here; I have the back one; and I have this side one.0309

So, I have four of them; that means that I have four triangles: 1, 2, 3, 4.0321

This is what my lateral area is going to be: the area of this, plus this, this, and that--all four together--is the lateral area (everything minus the base).0335

Well, I know that, if this right here is s for side, and then my slant height I am going to label as l,0346

then look at this triangle right here: that is the same as this triangle.0364

The slant height is this right here, and then this would be the side.0372

To find the area of this triangle, it is going to be 1/2 the base, which is s, times the l,0385

plus the same thing here: 1/2 the base of s and the height of l, plus 1/2 sl, plus 1/2sl.0396

And again, how did I get this? This is just 1/2 base times height; it is the area of a triangle.0410

But the base is a side, s; and the height is l, for the slant height.0417

So, all I did was to replace the base with s, and then the height with l; and that is the area of this triangle here, and the area of this triangle...0425

And so, I have four of them together; and this is all lateral area.0434

Now, if I factor out the 1/2 and the l from each of these, then from here, I am going to be left with s, plus, from here, s, plus s, plus s.0441

Well, all 4 s's together make up s + s + s; it makes up the perimeter of the base, so it will just be 1/2 times the slant height times the perimeter of the base.0468

And that is how we get this formula: 1/2 times the slant height times the perimeter of the base.0492

And to shorten it, you can just do 1/2, capital P for perimeter of the base, times l, which is the slant height (l is the label we are using for slant height).0501

This is lateral area; again, if you want to think of each of these triangles, we have four of them;0516

to find the area of each one, we know that, to find lateral area, we have to add the area of each triangle;0526

so I just wrote them out; I factored out 1/2 and the l; I am left with 4s, which is the perimeter of the base; and that gives you 1/2Pl,0533

1/2 times the perimeter of the base times the slant height; and that is the formula for the lateral area of a regular pyramid.0548

And then, to find the surface area of a pyramid, we know that it is the lateral area, plus the base.0561

There is only one base, and that would be that right there; so just find the area of that base.0570

And this is a regular pyramid, so it is going to be a square; so you find the area of a square.0577

We add it to the lateral area, and we know that lateral area, again, is 1/2 the perimeter of the base, times the slant height;0585

plus capital B for the area of the base is surface area.0598

Next is a cone: now, we know what that looks like--we have had ice-cream cones before.0611

It is a circular base, and then they meet at a vertex.0619

Now, the axis of a cone would be the segment from the vertex down to the center of the circle; that is the axis.0629

In a right cone, the axis and the altitude are the same; but when it comes to oblique cones,0641

because it is a little to the side, slanted to the side, leaning over, the altitude has to be perpendicular;0647

the axis, though, is still going to stay from the vertex of the cone down to the center of the circle.0657

So, the axis for the right cone is going to be the same as the altitude, but not for the oblique cone.0663

And we know that altitude is the perpendicular height.0676

Now, for the lateral area of a right cone, we are going to be measuring everything around (not including) the base (not the circle).0682

So, to find the lateral area, it is π times the radius, times the slant height.0695

So, slant height would be the measure of this right here; that is the slant height, l.0703

This right here is the altitude; that is the height of the cone; but going this way, that is the slant height.0710

Think of the height being slanted; that is l; we know that this is r; so the lateral area is π times r times l.0721

And then, of course, the surface would be just all of that, the lateral area, so π times r times the l,0744

plus the area of the base (capital B for the area of a base); and since it is a circle, we know it is πr2.0751

So, it is πrl + πr2.0763

Now, as long as you remember just this, you know that surface area is just the lateral area plus the area of the base.0768

So, this is what you actually really have to remember; and then, for surface area,0777

just remember that it is lateral area, plus the area of the base, the area of the circle.0784

Let's go over some examples: Determine whether the condition given is characteristic of a pyramid, prism, both, or neither.0792

Remember: a pyramid is when we have a polyhedron (because all of the sides are flat surfaces), and a vertex;0803

so if you have a base like this, it will go to each of these sides like that; that is a pyramid.0817

A prism is if I have something like that; that would be a prism, where we have two bases opposite and parallel and congruent.0824

So then, the first one: The lateral faces are triangles (let me highlight this and shade in that base).0851

The lateral faces of a prism all have to be rectangular; they are all rectangles--all of these lateral faces here.0862

For the pyramid, all of them have to be triangles; so this one here, we know, is "pyramid."0870

And the next one: There is exactly one base.0881

Well, prism, we know, has two bases; pyramid, we know, has only one; the opposite side of it would be the vertex; so this one is also "pyramid."0884

The next one: Find the lateral area of the regular polygon.0902

Lateral area is 1/2 times the perimeter times the slant height.0907

If you ever forget the formula, just think of it as finding the area of all four triangles.0919

If you just find the area of this triangle here, because it is a regular pyramid, each of the lateral faces is a triangle, and they are all congruent.0927

So, if you just find the area of one of the triangles, then you can just multiply that by 4, because I have 4 of them.0938

They are not always going to be 4; it depends on the base.0947

Find the lateral area: 1/2 the perimeter--if this is 12, then this is 12, and this is 12, and this is 12.0952

The perimeter is 12 times 4, which is 48; and then, the slant height...0963

now, be careful here: this is not the slant height; the slant height would be the height of the triangle, of the face, which is the triangle.0978

So, if that is one of our lateral faces, this is 12, and this is 10; then, slant height has to be that right there.0995

Be careful: this is not the slant height, so you have to look for it.1012

If the whole thing is 12, I know that this part right here is 6; so then, to solve for my (I'll just label that l) slant height,1018

it is going to be...using the Pythagorean theorem...l2 + 62 = 102.1031

This is l2 + 36 = 100; then l2 is going to be 64, which makes l 8;1040

so we know that the slant height, that right there, is 8.1055

And that is the measure that I need, 8; so the lateral area is going to be 48 times 8, divided by 2.1068

Or you can just do 24 (because I cut this in half), times 8; so on your calculator, 24 times 8 equals 192.1081

And I don't see any units, so it will just be units squared.1098

The next example: Find the surface area of the pyramid.1110

Surface area is the lateral area, plus the area of the base; the lateral area is 1/2 times the perimeter times the slant height,1114

and then, plus the area of the base...which is going to be side squared, because if this is a side, and this is a side...1133

we don't even need this; this is a side, and it is a square; so length times width is just side squared.1143

1/2...the perimeter would be...oh, do we know the side?1156

Well, we are not given this, but since we know that this is the center, and we are given half of that--1163

we are given from the center to this side, then if this half is 6, then the whole thing has to be 12, so this is 12.1173

So, the perimeter would be 12 times 4, which is 48.1184

The slant height...also be careful here: they give us the altitude--they give us the height--of this pyramid; but we don't know the slant height.1192

So again, we have to solve for it; now, slant height, we know, is from the vertex;1204

and just look at one of the triangles, one of the lateral faces, and then find the height of that.1214

If you are looking at a triangle, the lateral face, then it has to be that right there; that is the slant height.1222

OK, so then, here, how would we find the slant height?1237

Here we have a triangle; this is a right triangle; so this is 8, 6, and then the slant height would be the hypotenuse.1242

Using the Pythagorean theorem, 82 + 62 = the slant height, squared; as you can see, that is the right triangle.1254

This is 64 + 36 = l2; this gives you 100, which is the slant height squared;1264

and then, if you take the square root of that, then you get 10; so I know that this slant height is 10.1275

It is 10, plus...the area of the base is 12 times 12, which is 144; then, you just solve it out; so 24...1284

I just divided this by 2...times 10 is 240, plus 144 is going to be 384 units squared, and that is my surface area.1299

And the fourth example: Find the lateral area and the surface area of the cone.1330

My lateral area: the formula is π times the radius times the slant height--that is the lateral area.1339

I have the radius, and I have the slant height; surface area is the lateral area, plus the area of the base, which is the area of the circle.1354

First, let's look for the lateral area: LA is π times radius (is 5), and the slant height would be this right here.1373

So, on your calculator, you are going to multiply out π times 5 times 13; and I get 204 and 20 hundredths inches squared.1392

It is area, so make sure that it is units squared.1418

Then, to find surface area, let's first find the area of this base right here.1425

The area of the circle is B for area of the base; it is πr2; that is π, and then the radius is 5, squared, which gives me 78.6.1433

And we'll just do π times 5 squared; make sure that you square this first, and then multiply it.1453

Because of Order of Operations, you have to do the exponent before you multiply: 25 times π is 78.54, or 78.6;1464

so, my surface area is going to be my lateral area, that number right there, added to my area of the circle.1478

And then, when you add it together, we should get 282.74; and then here, it is inches squared.1498

Now, if you remember from the prism, remember: when you find surface area, you are not just adding the base.1516

If you have two bases, then you have to multiply this number, the area of the base, times 2, because you have 2 of them.1527

You have to take that into consideration.1533

I only have one base, so whatever my lateral area is, I can just add this number to that.1536

But if I have two bases, then I have to multiply that base times 2 and then add it.1543

Just remember: when you find surface area, it has to cover every single part of your solid--every single side, lateral faces and your bases.1552

That is it for this lesson; thank you for watching Educator.com.1567

Welcome back to Educator.com.0000

For the next lesson, we are going to go over volume of prisms and cylinders.0002

Volume is the measure of all of the space inside the solid.0011

We went over lateral area and surface area: remember, lateral area measures the area of all of the outside sides, except for the bases;0020

and then, surface area would be the area of all of the sides together, including the bases.0035

If you were to wrap a box, let's say, with wrapping paper, that would be surface area.0044

If you were to fill the box with something (let's say water or sand or anything--just filling it up), that would be volume.0051

That is going to be the volume of that box.0059

Now, the box itself is a prism; prisms, remember, are solids with two congruent and parallel bases.0065

They have to be opposite sides; they are two congruent and opposite bases.0078

It can be any type of shape; it can be a triangle; it can be a rectangle; it can be a hexagon, pentagon...whatever it is.0082

And the rest of the sides, the lateral sides (meaning the sides that are not bases) have to be rectangles; that is a prism.0091

This right here is a rectangular prism, because the bases are rectangles.0102

Let's say that we are going to name that the base; then, that means that the bottom is also the base.0111

The volume (meaning everything inside, the space inside) is going to be the area of the base, times the height--0125

the area of this right here, times the height of the prism.0140

When it comes to a rectangular prism, we know that the area of the base is the length times the width.0147

And then, the volume will be times the height; so length times width times height is for a rectangular prism.0157

Length times width measures the area of the base; any time you have a lowercase b, that just means the segment base--0168

the line segment, like maybe the measure of the bottom side or something; that is the base, lowercase b.0178

When you have a capital B, that is talking about the area of the actual base, the side base.0186

A capital B is the area of the base, times the height; and this is the volume.0194

For rectangular prisms, it would just be the length times the width; but for any type of prism, it is going to be capital B,0201

for base, the area of the base, times height; so this is the formula for the volume of a prism.0210

Now, the volume of a cylinder: remember: a cylinder is like a prism in that there are two bases, opposite and congruent and parallel.0220

But for a cylinder, the bases are circles: circle and circle.0232

Now, to find the volume, again, we are measuring the space inside.0241

So, if you were to take a can, and you were to fill it up with water or something, that would be volume.0247

How much water can that can take?--that would be volume.0252

The formula for this is actually the same as the formula for the prism: πr2 is the area of the base;0263

that is the formula for the area of a circle (which is the base), times the height.0277

So, for a cylinder, volume is also capital B, for the area of the base, times the height.0285

So again, prism and cylinder have the same formula for volume.0295

Just think of capital B as the area of the base; whatever the base is, it is the area of that side, times the height.0300

The first example: we are going to find the volume of this prism.0313

Now, remember: when you have a prism, all of the lateral sides--meaning all sides that are not bases--have to be rectangular.0317

If you have a prism, and you know that it is a prism for sure, then the sides that are not rectangular automatically become the bases.0326

So here, I see a triangle; that would have to be the base, the side that is the base,0336

which means that the side opposite has to also be the same; it has to also be congruent; if it is not, then it is not a prism.0346

There is our base; to find the volume, remember, you are going to find the area of that base, times the height.0358

The area of this triangle, we know, is 1/2 base times height (and this is lowercase b, times the height)--is all volume.0370

This base and the height are 6 and 10; we know that those two are our measures, because they are perpendicular.0388

The base and the height for the triangle have to be perpendicular, which they are; that is what this means, right here.0395

So, it is 6 and 10; so 1/2 times 6 times 10, times the height of the prism (that is 4)...we just multiply all of those numbers together, and that is our volume.0400

Here, 1/2 times 6 is 3, so I just cross-cancel this number out: 3 times 10 is...so all of this together is going to be...30,0417

times the 4, is 120; our units are inches...for volume, it is going to be units cubed.0433

For area, we know that it is squared; volume is always going to be cubed.0447

Here, this is the volume of this prism; again, it is the area of the base, 1/2 base times height, times the height of the prism.0454

Be careful that you don't confuse this h with this h; this h is just the height for the base, for the triangle;0466

and this height is the height of the actual prism.0477

The next example: Find the volume of the cylinder.0485

Again, the formula is the same: capital B (for area of the base), times the height.0490

In this case, the base is a circle; the area of a circle is π (oh, that is not π) r2, times the height.0500

πr...the radius is 4, squared, times 10 for my height...just to solve this out, to simplify this,0519

it is going to be 42, which is 16, times 10; that is 160π.0533

To turn that into a decimal, to actually multiply this out, 160 times π, you can go ahead and use your calculator.0541

I have mine on my screen; so it is 160 times π; so my volume becomes 502.65; my units are centimeters, and then volume is cubed.0548

So, there is the volume of my cylinder.0570

The next example: A regular hexagonal prism has a length of 20 feet, and a base with sides 4 feet long; find the volume of the prism.0577

Let's see, a regular hexagonal prism: "hexagon" means 6 sides, so that means the base of my prism is going to be a hexagon.0591

Let me try drawing this as best I can--something like that.0604

It has a length of 20 feet and a base (that is this hexagon right here) with sides four feet long; so each of these is four feet.0623

Make sure that you remember that it is regular, so it has to be equilateral and equiangular.0635

That means that each of these sides is going to be 4 feet long; find the volume of this prism.0640

Back to the formula: volume equals the area of the base, times height.0648

Now, to find the area of this base, it is a hexagon; so remember: to find the area of a regular polygon...let me just review over that quickly.0658

If I have a regular hexagon (that doesn't really look regular, but let's just say it is), it is as if I take this hexagon,0673

and I break it up into congruent triangles: I have 1, 2, 3, 4, 5, 6 triangles.0682

The area of each of these triangles is 1/2 base times height; so if this is the base, and this is the height (that is for this triangle),0690

and then I have 6 of them--so then, here is 1, 2, 3, 4, 5, 6--it is 1/2 base times height; multiply that by 6.0705

Well, let me just use a different color, just to show you that this is on-the-side review.0719

If I take the base (this is the base), and I multiply it by the 6, well, if this is the base, there are 6 sides.0725

Then, my base and my 6 become the perimeter; and then, my height of the triangle, this right here, is what is called the apothem.0739

And then, my 1/2...so then, the area of this whole thing, which is 1/2 base times height, times 6,0758

because there are 6 triangles, becomes 1/2 times the perimeter times the apothem.0766

So, this base is 1/2 times the perimeter times the apothem, and then times this height right here, the height of the prism.0774

Let's see, my perimeter is going to be (since it is a regular hexagon) four times...I have 6 sides, so...4 times 6 is 24;0786

and then, my apothem, remember, is from the center to the midpoint of one of the sides; so that is perpendicular.0809

That is my apothem; now, I don't know what my apothem is, so I have to solve for it.0825

So, I am going to make a right triangle; a right triangle is always the best way to find unknown measures.0831

I am just going to draw that triangle a little bigger, just so that you can see.0845

My apothem (that is this right here): if this whole side is 4, then I know that this right here has to be 2; it is half.0854

And then, since that is all I have to work with, let's see: if I had this measure,0866

I could use the Pythagorean theorem, a2 + b2 = c2; but I don't.0872

If you did have that, then you could use the Pythagorean theorem.0878

What I can do: I know that from here all the way around, passing through all 6 triangles, it is going to have a measure of 360 degrees.0882

I just want to find the measure of this angle right here: 360 divided by each of the 6 triangles is going to be 60 degrees.0901

That means that this right here, just one triangle, is going to have a measure of 60;0914

it is going to have a measure of 60; this is 60; 60; and 60, and so on.0923

If from here, all of this is 60, then I know that half of this right here has to be 30 degrees.0930

If this is 30, and this is 90, then this has to be 60; and that is 0...60 degrees.0941

So then, this becomes a 30-60-90 degree triangle; and again, I found out that this is 300950

by taking my 360, which is the whole thing, and dividing it by 6 (because there are 6 triangles).0958

So then, this triangle and this triangle, all 6 of them--each of them is going to have an angle measure of 60 degrees.0968

So, this is 60; but then I have to cut it in half again, so then, that is 30.0975

This becomes a special right triangle--remember special right triangles?--it is a 30-60-90 degree triangle.0981

The side opposite the 30 degrees is going to be n; the side opposite the 60 is n√3; the side opposite the 90 is 2n.0992

n is just the variable for the special right triangles; that is what I am going to use.1004

Now, what side is given to me? I have a 2 here--that is the side opposite the 30.1011

That is this right here; so I am going to make those two equal to each other: n = 2.1016

Well, if n equals 2, then what is the side opposite the 60? That is going to be 2√3.1024

Then, the side opposite the 90 is going to be 2 times n, which is 4.1029

OK, that means that the side opposite the 60 is 2√3; the side opposite the 90 is 4.1035

My apothem, which is this side right here, is going to be 2√3; my height of the prism (because we are back to this volume formula--1044

not the height of the triangle, but the height of the prism) is 20 feet.1063

So again, it is 1/2 times the perimeter, times the apothem, times the height.1071

To find the perimeter, remember (and all of this, 1/2Pa--that is for the area of the base): you are going to take the 4,1078

and multiply it by all of the sides; that is 24; to find the apothem, you just use special right triangles.1087

This is 2, so then this is 2√3, and then multiply that by 20; so I am going to have...1097

and then, you can just use your calculator for that...times that number, times...and I get 831.38 feet cubed for volume.1109

So, the volume of this prism is that right there.1139

And then, the fourth example: Find the volume of the solid.1147

For this, we have two prisms stacked like that; so to find the volume of this whole thing, I need to find the volume of this prism,1152

the one on the bottom; find the volume of this top prism; and then add them together.1165

I am going to say that this is prism #1 here; this is #1, and this is #2.1175

#1 is going to be...volume equals capital B (for the area of the base), times the height.1185

The base...this is a rectangular prism, so it is 10 times 10; the length times the width is 10 times 101196

(because it is the same; they are congruent) times the height, which is 4.1214

So, the volume of this first prism, prism #1, is 100 times 4, which is 400 meters cubed.1219

Then, I have to find the volume of prism #2; it is also a rectangular prism, so length times width times the height.1231

The length and the width are congruent, so this is going to be 6 times 6; and the height is 2.1248

I know that 6 times 6 is 36, times 2 is 72; that is meters cubed.1261

The volume of this solid is going to be prism 1, plus prism 2.1273

Prism 1 is 400 meters cubed, plus the 72 meters cubed, which is going to be 472 meters cubed.1280

Make sure that you add the two volumes together; you are not multiplying them,1296

because it is the volume of that whole thing, plus the volume of this whole thing.1303

You are adding them together--the total is the sum of them.1309

That is it for this lesson; thank you for watching Educator.com.1316

Welcome back to Educator.com.0000

For the next lesson, we are going to go over volume of pyramids and cones.0002

Remember that volume is how much space the solid is taking up--how much space is inside the actual solid.0012

To find the volume of a cone, it is 1/3 times the area of the base, times the height.0018

Remember that, for the area of the base, we are going to use capital B; so whenever you see capital B, that is for the area of the base.0027

Lowercase b is just going to be the segment base; capital B is for the area of the actual base of the solid.0037

The volume of a cone is 1/3 the area of the base times the height.0046

Now, let's say that this cone has a radius of 3 centimeters, and has a height of 5 centimeters.0055

To find the area of the base for that, because the base of a cone is a circle, we know that the area of the base is πr2.0076

This is 1/3 times πr2 times the height: 1/3 times π times 32 times 5 for the height.0115

This is 9π times 5; I can go ahead and divide this number by 3, because this is over 1; so then, this simplifies to 3 times π times the 5.0139

So, this will be 15π; now, to simplify this out, you can just use your calculator (I have a calculator here on my screen):0141

it is going to be 15 times π is 47.12; the units will be in centimeters, and for volume, it is always cubed, remember.0152

So, this cone has a volume of 47.12 centimeters cubed; that is the volume of a cone, 1/3 times the area of the base times the height.0169

For a pyramid, it is actually the same exact formula: it is 1/3 times the area of the base times the height.0184

Now, if you were to base the formula on what kind of base of the solid you have, then it would have a different formula.0190

But because we can just say that it is every base, no matter what the base is,0201

no matter what type of polygon you have as the base, it will always be 1/3 area of the base, times the height.0207

Base for this, because we have a rectangle, is going to be the length times the width, and then times the height.0221

So, let's say I have...this is 6; this is 5; and my height is 10, and let's say this is all in meters.0233

Then, it is going to be the length, which is 6, times the width of 5, times the height of 10.0254

Now again, you can just go ahead and divide this and simplify this: 6 divided by 3 is 2, so it would be 2, times the 5, times 10.0265

And that is 10 times 10, is 100; my units are meters cubed.0278

And that is how you find the volume of the cone and the pyramid: 1/3 the area of the base, times the height.0288

For our first example, we have a pyramid, and the base of this pyramid0297

is a regular pentagon with the area of 20 kilometers squared; find the volume of the pyramid.0303

So, this base, the pentagon, is 20 kilometers squared, and we have to find the volume; so then, the volume is 1/3 area of the base times the height.0311

Now, we don't have to solve for the area of the base, because it is already given to us: so it is 1/3 times 20 times the height of 9.0327

Let's divide 9 and 3; that is 3, times 20, which is 60 kilometers cubed; that is the volume of this pyramid.0342

OK, for the next example, find the volume of the solid: here we have a cylinder, and we have a cone on top of the cylinder.0362

We are going to have to find the volume of each separately, and then we are going to add them together.0374

Now, if you notice, this solid right here, this cone, is not a right cone, because the vertex and the center of the base are not perpendicular.0379

This is called an oblique cone; to find the volume of this cone, it is actually the same exact formula,0393

but you just have to be careful with the height, because the height is no longer going to be from the vertex to the center of the base.0411

Just make sure that you remember that, for the height, it has to go from the vertex down to a point on the base, perpendicularly.0420

So then, this would be the height of this oblique cone.0431

Now, to find the volume of a cylinder, remember: it is going to be the area of the base, times the height.0435

For both solids, the base is going to be a circle: the radius is 5; 1/3πr2 for the base...5 squared...the height is 7;0448

so, this is for the cone, and then this right here is for the cylinder: volume is, again, πr2,0470

so that is π(5)2, times the height of 15.0490

When I add them together, I can just say that the volume of the solid is then going to be 1/3 times 25π0500

(I just squared this) times the 7, plus, again, 25π, times 15.0510

So here, this is the volume of the cone, and this is the volume of the cylinder.0524

I am going to use my calculator: I'll just punch it in...divide by 3...so the volume of the cone is 183 and 26 hundredths,0534

plus the next one: 1178.10; and then, when I add them together, cone plus the cylinder, I get 1361;0555

and the units are in centimeters, and volume...so it is cubed.0604

Make sure, when you have some kind of shape like this, that you recognize that this is a cone; this is a cylinder;0614

find the volume of each, and add them together, and that is going to be the volume of the whole solid.0620

A third example: Find the volume of the pyramid: here our base is a triangle; this is our height.0629

Now, what do we have here? To find the area of the base (because, again, my formula for volume0641

of this solid is going to be 1/3 the area of the base times the height), I am going to take this triangle;0652

I am going to re-draw it so that it is easier to see: this is 11 meters, this is 11, and this is 5.0668

Now, we don't have the height here, but I know that this is an isosceles triangle, so if I do that,0681

I can use half of this to make it a right triangle, and to use the Pythagorean theorem: so this is going to be 2.5; here is 11;0694

I am going to use those two to find this unknown.0704

So, if I call this a, it is going to be a2 + 2.52 = 112.0709

And this is the Pythagorean theorem: a2 + b2 equals hypotenuse squared.0719

Just use your calculator, and get a2...you square this, and you square this;0731

you are going to subtract this from this, and I get this; and then, you are going to take the square root of that, and a is around 10.7.0743

So again, I had to find this because, to find the area of this base right here, I need the height.0773

I don't have the height, so to find it using the Pythagorean theorem, I am going to do a2 + b2 = c2.0782

And then, that is going to be 10.7 for the height.0792

Then, the area of the base is going to be 1/2 base, which is 5, times height, which is 10.7 (what we just found).0799

And then, we get 26.something if we round that number; so then, here I am just going to plug this into the formula: 26.20.0817

And then, the height--do I have the height of the solid?0837

Let's see, this is my height of the solid; I don't have it, so I have to look for it.0842

Now, again, since this is 11, and this is 16, we can use this triangle (it is a right triangle), and using the Pythagorean theorem, I can find this height.0849

So, I am going to say h2 + 112 = 162, the hypotenuse squared.0862

My height squared is 135; take the square root of that; h is 11.62.0884

I am going to include that into my formula: 11.62, and then just go ahead and use your calculator for the rest of that.0898

Multiply it by 6.78, and divide that number by 3; my volume is going to be 103.72; my units are meters cubed.0907

So again, this one took a little extra work, because we were missing some measures.0931

The height wasn't given to us; we had to find this height, so we used the Pythagorean theorem to find that height.0937

And then, to find the area of the base, this triangle, since the area of a triangle is 1/2 base times height, we had to find the height.0944

And we did that by the Pythagorean theorem: cutting this side in half, since it is an isosceles triangle, and then using this.0956

We found the area of the triangle; and once you find all of those unknown sides, just go ahead and plug them into the formula.0972

And use your calculator to solve it out.0979

For the fourth example, we are going to find the volume of the octahedron.0984

Now, we know that an octahedron is a regular polyhedron with 8 sides.0989

So, I have four sides up here, and I have four sides for this bottom part.1001

Let's break this up into two solids, two pyramids.1008

Now, because it is a regular solid, I know that all sides, all of the edges, have a measure of 10 centimeters.1011

This right here is the slant height; it is not the height of the actual solid.1021

So again, let's break this up into two pyramids, and then we are going to just add them together1027

(or multiply them: if we find the volume of one of the pyramids, let's say the top half--1035

the top half and the bottom half are exactly the same, so we could take this volume and then just multiply it by 2 to find the whole thing).1040

Again, I am just going to find the volume of the top half, the pyramid of the top part of it.1055

So then, here is the base; it is regular, so all of the sides are going to have a measure of 10 centimeters.1061

And then, something like that...I'll just do this one over...1071

Now, I need to find the actual height of the solid, from here all the way down to here, in the center.1091

I know that (and this is the slant height) the slant height has a measure of 8.7.1103

This base right here is 10; this is 8.7; let me just draw these sides blue, so that you know that it is this right here,1115

with this a side, the red side, the slant height; so this is 8.7; this is 10, because all of the sides are 10.1135

Then, half of this is going to be 5 centimeters long.1146

So then, using the Pythagorean theorem, this became 8.7; this is 5, because it is half of a side; so this is 5; this is 8.7.1152

Then, what is this right here--isn't it also 10?1169

So then, the height of that is going to be 10, because in a triangle, if we have 5 and then 8.7, then the hypotenuse is 10.1172

So, it is all going to be the same; the height is also 10.1184

The volume of this pyramid is 1/3 the area of the base, times the height.1191

So, it is 1/3...area of the base is 10 times 10, because it is a square...and then the height is 10.1201

Now, again, I am taking this whole thing and cutting it in half to find the volume of the upper part.1222

I am now going to take all of this, whatever that is, the volume, and then multiply it by 2 to find the area of the whole thing.1229

The volume of the solid (this is the volume formula) is going to be 1/3 times the area of the base, times the height,1240

and then times 2, because it is as if I am taking this whole volume, and then adding it to this volume;1259

and that is just times 2, because they are the same.1268

So, 10 times 10 is 100, times 10 is 1000; 1/3 times 1000 times 2 is 2000, divided by 3;1271

and then, using your calculator, the volume is 666.67 centimeters cubed.1295

And that is it for this lesson; thank you for watching Educator.com.1319

Welcome back to Educator.com.0000

For the next lesson, we are going to go over surface area and volume of spheres.0002

First, let's go over some special segments within the sphere.0009

A radius, we know, is a segment whose endpoints are on the center and on the sphere--anywhere from center0014

(it has to be here), and then anywhere on the sphere--this is the radius.0027

A chord is a segment whose endpoints are on the sphere; it could be anywhere on the sphere.0032

So, it could be from here all the way to here; this is a chord, and this would be a chord; this would be a chord.0040

A diameter, we know, is a segment whose endpoints are on the sphere, but it has to pass through the center.0057

This is going to be a point on the diameter; it can go from here (that is the back side of the sphere),0065

and then passing through the center, and then to another point on the sphere; that would be the diameter.0074

And then, a tangent is a line that intersects the sphere at one point.0081

A chord, we know, is at two points; but this intersects only at one point.0088

So, it would be like this, where it is just intersecting at one point; that would be a tangent.0093

When it comes to intersecting a plane, a sphere can intersect at one point.0107

This is the plane, and here is a sphere; it is intersecting at one point.0114

Here, it is intersecting at a circle; so we can see the intersection where the plane and the sphere meet--it is this little circle right there.0119

That is where they are cutting.0131

And then, here, this is also a circle; but this is called a great circle, because it is the biggest circle0134

that can be formed from the intersection of a plane and a sphere.0146

It is basically passing through, intersecting, at the center of the sphere; and that is called the great circle.0150

This circle right here is called the great circle; this is just a circle, because the center is somewhere right here; it is not passing through the center.0157

But because this is passing through the center, it is the largest circle that can be formed with an intersection.0168

Now, with that intersection, this circle, the great circle, cuts the sphere in half; and that is called a hemisphere.0177

So, a half of a sphere, separated by a great circle, right at that great circle (half of the sphere)--this is a hemisphere.0188

To find the volume of this, you have to just divide it by 2.0200

OK, and then, the surface area of a sphere is 4πr2.0206

This is the radius; it is just the area of the circle, times 4; it is 4πr2, and that is surface area.0217

Remember: surface area is just the area of the whole outer part.0233

It is not the space inside (that would be volume); it is just the outside of it--that is the area, 4πr2.0237

And then, to find the volume of the sphere, it will be 4/3πr3.0249

The surface area is 4πr2, and the volume is 4/3πr3.0256

Let's go over our examples: Determine whether each statement is true or false.0266

Of a sphere, all of the chords are congruent.0271

We know that that is not true, because, within the sphere, I can have a chord passing just from here to here;0275

that is a chord (as long as the endpoints are on the sphere); I can have it passing through from here,0289

all the way to here; that would be a chord; so, chords are not all congruent--they could be, but they are not all congruent.0298

So, this would be a false statement.0308

In a sphere, all of the radii are congruent.0313

We know that this is true, because, whether the radius is going from here to here, or from here to here,0319

from here...because they are always from the center to a point on the sphere, these are all congruent; that is true.0328

The longest chord will always pass through the sphere's center.0342

Now, in order for it to be a chord, it just has to have the endpoints on the sphere.0347

If it passes through the center, we know that that is called the diameter; but that is also a chord, because the endpoints are on the sphere.0352

A diameter is a special type of chord; so this is true, because we can't draw a longer chord that doesn't pass through the center.0364

The next example: we are going to find the surface area of the sphere.0378

Here, again, for surface area, we are finding the area of the outside part--how much material is being used on the outside.0384

The surface area is 4πr2, 4 times π...the radius is 5...squared; so this is going to be 4π times 25.0396

4 times 25 is 100, so it is 100π; or because we know that π is 3.14, 100 times that will be 314.0417

And the units are inches...this is surface area, so it is squared; the surface area of the sphere is 314 inches squared.0431

OK, the next example: Find the volume of a sphere with a diameter of 20 meters.0447

We know that a diameter is twice as long as the radius--from here, all the way going through, is the diameter;0456

and that is 20; so the radius is going to be half of that, which is 10 meters.0463

To find the volume, it is 4/3πr3, so the radius is 10, cubed...4/3π...10 cubed is not 30;0470

it is 10 times 10 times 10, which is 1000; so if I multiply 4 times 1000, is going to be 4000 times the π, over 3.0494

And you can just simplify that with your calculator: it is 4000 times π, divided by 3; and for the volume, I get 4188.79.0513

My units are meters...be careful here; it is not squared, because we are not dealing with area;0536

we are dealing with volume, how much space is inside this sphere; and that is cubed.0541

Now, the fourth example: we are going to find the surface area and the volume of the solid.0559

Here, we have a cylinder, and then let me just finish this off here like that, just so that you can see the cylinder.0565

And we have part of a sphere; now, let's see if this is a hemisphere (a hemisphere, remember, is half a sphere).0575

Here, the radius is 6, and this is also the radius of 6; we know that that is the radius, because that is the same measure.0589

So then, this has to be a hemisphere; now, be careful--just because you see part of a sphere doesn't automatically mean that it is a hemisphere.0597

It doesn't mean that it is half the sphere; here, I had to make sure, because,0605

from the center all the way from each side (to the point on the sphere), the radius has to be the same; then, that makes this a hemisphere.0613

So, to find the volume of this hemisphere (h for hemisphere), it is going to be the volume of a sphere, 4/3πr3;0625

and then, I am going to multiply this by 1/2, or divide the whole thing by 2, because it is half of a sphere.0645

That would be the volume of a hemisphere: just the volume of the whole thing, divided by 2.0656

And then, I have to find the volume of this cylinder, because it is the cylinder with the hemisphere.0664

So, when I find the volume of the cylinder (I'll write c for cylinder), that is the area of the base, πr2, times the height, which is 8 there.0671

There is the formula for the hemisphere and the cylinder; and then I have to add them together.0689

Let's first find the volume of the hemisphere: 4/3π...the radius is 6, cubed; and then multiply that by 1/2.0698

OK, I am going to go ahead and multiply these fractions together: it is 4...and you can just go ahead0714

and start punching it into your calculator if you want to...here, this is 4, over 3 times 2 is 6, π, times 6 cubed.0720

Times the π...times the 4...and divide that by this...if it makes it easier for you, you can just find the volume0742

of this whole sphere first, and then just divide that by 2; I just went ahead and multiplied the fractions.0756

It doesn't really matter for now; just make sure that you cube this before multiplying; you can't multiply all of this with the 6,0764

and then cube it, because with Order of Operations, remember: you have to always do an exponent before you multiply.0775

I get 452.39; this is the volume of the hemisphere; units are cubed for volume.0783

And then, I have to find the volume of the cylinder, π...my radius is 6, squared, times the 8.0801

It is π, times 36, times 8; on my calculator, I do π times 36 times 8, and I get 904.78.0819

So then, I take this; I have to add it to my cylinder; add those two numbers together, and I get 1357.17; this is centimeters cubed.0848

The whole thing is centimeters cubed; there is my answer for the volume of this whole solid.0874

That is it for this lesson; thank you for watching Educator.com.0884

Welcome back to Educator.com.0000

For the next lesson, we are going to go over congruent and similar solids.0001

Whenever we have two solids that are either similar or congruent, there is a scale factor.0010

A scale factor is just the ratio that compares the two solids; it is the ratio of the corresponding measures (it has to be corresponding).0018

If we are going to use this side for the scale factor, then we have to use the corresponding side of the other solid.0028

So then again, the scale factor is the ratio of the two similar solids.0039

Here, the scale factor...since this is 2 and this is 4, we are going to say that it is 2:4; simplified, this is 1:2.0046

The scale factor is 1:2, or we can say 1 to 2, like that; it is just the ratio between the two similar solids.0060

For congruent solids, these have to be true: the corresponding angles are congruent; the corresponding edges are congruent0072

(we have to have congruency between the two solids); the areas of the faces are congruent; and the volumes have to be congruent.0082

And congruent solids have the same size and same shape.0094

Remember that, for congruent solids, it is same size and same shape; for similar solids, it is going to be different sizes, but same shape.0100

Remember: whenever we have something similar, it has to be the same exact shape, but then just a different size.0113

Congruent solids will have the same shape and the same size.0120

And the scale factor is going to be 1:1, because obviously, the corresponding sides and the corresponding parts are going to be the same.0124

So, it is going to be a ratio of 1:1.0132

Looking at similar solids, if the scale factor is a:b, then the ratio of the surface areas is going to be a2:b2,0139

and the ratio of the volumes is going to be a3:b3.0153

Let's say we have two solids, and the scale factor between the two is 2:3;0161

then the ratio of the surface is going to be 22:32, so it is going to be 4:9.0168

If that is the scale factor (that is the ratio between the two solids), their surface areas are going to be 4:9.0184

And then, the ratio of the volumes is going to be 23:33; 2 cubed is 8; 3 cubed is 27.0194

That is going to be the ratio of their volumes.0209

The first example is to determine whether each pair is similar, congruent, or neither.0218

Looking at these two, this pair: here, this is a cube, because we know that all of the sides are going to be congruent.0225

So, this is a cube; this is also a cube with all of the edges measuring 5.0241

So, in this case, because they are the same shape, but just different sizes, this is similar.0248

And we always know that all cubes are going to be similar, because cubes have the same shape.0258

No matter how big or how small, all cubes are the same shape; they can just be different sizes.0267

If they are the same shape, but same size, then they would be congruent; if all of these were also 8 inches, then they would be congruent.0273

But since they just have the same shape and different sizes, they are just similar.0282

And then, these two: let's see, here we have 24; that is diameter; from here to here is 26; we don't know the height.0289

Remember: for these two to be congruent, they have to have congruent corresponding parts.0303

To find the height here (because I don't know the height), I know the height here; this would be the height for this one,0315

because it is just the cylinder that is turned sideways; so if we say that that is the height,0320

then I need to find the height of this, so that I can compare.0327

The diameter is 24 here; the radius is 12 there; so the radius here will also be 12, because the diameter is twice the length of the radius.0331

To find the height here, I am going to use this triangle; and this is a right triangle, so then I can just use the Pythagorean theorem.0344

It is going to be...if I name that h...h2 + 242 is going to equal the hypotenuse (26) squared.0352

So, 24 squared is 576; and then, 26 squared is 676; so, if I subtract them, I am going to get 100, which makes my height 10 centimeters.0365

This is 10; this is also 10; so then, their heights are congruent; their radius is congruent.0393

So, if I were to find the area of the base, then it is going to be π times 12 squared (the radius is 12, so it is 12 squared).0402

Here, it is also going to be π times 12 squared; so the area of the bases will be the same.0414

To find the volume, it is going to be the area of the base...that is π, r, squared, times the height.0422

The same thing happens here: the radius is the same, and the height is the same, and we know that π is always the same.0432

So then, their volumes are going to be exactly the same.0444

Well, if we have two solids with the same exact volume, same shape, same size, same corresponding parts, we know that this has to be congruent.0447

And again, because they are going to have the same volume, they are congruent.0460

Determine if each statement is true or false: All spheres are similar.0469

Spheres always have the same shape; no matter what, all spheres are the same shape.0477

Now, sizes could vary; we could have a large sphere; we could have a small sphere.0483

But they are always going to be similar, because they always have the same shape.0488

So, any time two solids have the same shape, they are always going to be similar; so this is true.0492

The next one: If two pyramids have square bases, then they must be similar.0504

Well, even if they have a square base, yes, squares are always similar, because squares always have the same shape.0511

A square is a square, whether it is large or small; they are always going to be similar.0524

But for pyramids, we can have a tall pyramid, or we can have a short pyramid.0531

So, it doesn't always mean that they are going to be similar--these do not have the same shape.0550

So, even if their square bases are exactly the same (they are congruent), because we don't know the height, this is false.0557

If two solids are congruent, then their volumes are equal.0572

Well, let's say that we have exactly the same rectangular prism; it is congruent to this.0578

If they are exactly the same, then isn't the space inside also going to be the same?0592

So, all of this is going to be the same as all of this; so this is true.0601

Congruent solids have congruent volumes, the same volume.0608

For this example, we are going to find the scale factor and the ratio of the surface areas and the volume.0617

The scale factor between this and this prism that are similar is going to be 4:6.0624

We are going to use corresponding parts to determine the scale factor; it is going to be 4:6.0634

We need to simplify this, and it is going to be 2:3; that is the scale factor between these two prisms.0641

Then, to find the ratio of the surface area, for surface area, it is a2:b2.0649

And then, for volume, it is going to be a3:b3.0660

For surface area, the ratio is going to be 2 squared to 3 squared, which is 4 to 9.0668

Now, it doesn't mean that the surface area of this is 4 and the surface area of this is 9; it is just the ratio between this one and this one.0684

So, when we find the surface area, if we were to find the surface area of both this prism and this prism,0695

and then we simplify it, it is going to become 4:9.0701

And then, for volume, it will be 23 to...what is that one, 3?...33, so it is going to be 8:27.0710

And again, that does not mean that the volume of this is 8, and that the volume of this is 27.0727

Let's actually find the volume, given two corresponding sides right here.0736

This is similar, so the ratio between these two prisms is 3:2.0745

And make sure that you keep the ratio the same; if you are going to keep it at 3:2, that means that you are listing out this one first.0755

You are naming this first; so it is this one to that one.0761

If you want to go the other way, that is fine; but then, you are going to have to make the scale factor 2:3, instead of 3:2.0765

Always keep in mind which one this is: this number refers to the larger prism.0771

Now, the volume of the smaller one, the second one, is given; it is 50 inches cubed.0779

So, to find the ratio of the volumes, it is 3 cubed to 2 cubed; that is 27 to 8; that is the ratio of the volumes.0785

This is the larger one, over the smaller one; that means that, if I want to find the actual volume,0802

the volume of this one to the volume of this is going to become 27/8, simplified.0810

Then, I just know that I can make a proportion: this ratio is going to equal the volume of that0820

(because that one applies to the larger one), so let's say V for volume, over...what is the volume of this smaller one? 50.0827

That is how I make my proportion, because the volume of the larger to the volume of the smaller, simplified, is going to become 27/8.0838

Use the volume of the larger over the actual volume of the smaller prism.0848

So then, here I am going to solve out this proportion; this becomes 8V (cross-products: 8 times V) equals 27 times 50.0853

Using your calculator, 27 times 50 equals 1350; divide the 8; your volume is 168.75.0866

And then here, our units are inches (and for volume, it has to be) cubed--units cubed.0887

And that would be the volume of this larger prism.0897

OK, so again, to make your proportion, we know that this ratio has to equal this ratio.0901

They both are the ratios of their volumes; you have the volume of the larger prism to the volume of the smaller prism;0909

that is going to become 27/8; so this is simplified, but then their volumes have to equal 27/8; that is the ratio of the prisms.0923

You just make the two ratios equal to each other; set it equal.0934

Make sure that you keep the larger prism as your numerator; so it has to be 27 over the smaller prism.0939

And this is going to be V/50, the larger over the smaller.0949

If you do it the other way, if you do the smaller over the larger, then you have to make sure that you flip this one, also: 50/V.0953

Find cross-products, and then just solve it out.0960

That is it for this lesson; thank you for watching Educator.com.0962

Welcome back to Educator.com.0000

For the next lesson, we are going to go over mappings.0002

Mappings are transformations of a pre-image to another congruent or similar image.0006

When you have an image, and it has either a congruent or a similar relationship with another image,0014

then that is transformations, which is also mappings.0023

The different types of transformations are rotation, translation, reflection, and dilation--four types of transformations.0030

The first one, rotation, is when you take the pre-image (the pre-image is the original, the initial, the first image), and it rotates.0041

So, you turn it to make the second image; it is just rotating or turning.0055

Translation is when you take the image and you slide it, so it just moves; that is it.0063

It doesn't rotate; it doesn't do anything but just move--a slide or a glide.0070

Reflection is when you flip the image: you have two images, and they are just reflections of each other.0079

And dilation is when you enlarge or reduce the image.0086

Again, transformations are when you perform one of these four to a pre-image to create another image that is either congruent or similar.0092

Here are just some image: with rotation, you take this image (this is the pre-image), and to make this image, all I did was rotated it--just turned.0107

It is the same image, and it just rotated.0121

Translation: again, this is the pre-image, and it just slides or glides--just moves.0124

It stays the same; it just moves to a different location, a different place right here.0132

Reflection is, again, like a mirror reflection; they are reflections of each other.0138

And dilation is when an image gets larger or smaller; this is the same shape, but different size.0146

It just gets bigger, or it gets smaller; but it has to be the same shape.0157

And if two images have the same shape, but a different size, then we know that they are similar.0161

With dilation, it will be similar images; so then, the other three (rotation, translation, and reflection) are all congruence transformations,0167

because when you perform these transformations, they don't change; they are still congruent in size and shape.0180

Nothing changes; it is just the way you position it, or the way you rotate or reflect or translate the image; it is just going to stay the same.0188

And that is called an isometry; an isometry is a transformation that maps every segment to a congruent segment.0200

Again, when you either rotate, translate, or reflect, the images are congruent; they are the same.0209

Describe the transformation that occurred in the mappings.0220

Here, we want to know what happened with this image to get this image.0223

All that this did was to turn; so from this to this, it just turned a certain angle amount; and so, this is rotation, because it just rotated.0233

This right here, from this image to this image, the pre-image to the image, looks like a reflection; it looks like it is looking in a mirror.0253

It reflects, so this is reflection.0264

And then, for these two, see how one is bigger than the other.0271

So, even though it kind of looks like reflection, it can't be, because reflection has to be exactly the same.0276

It has to line up exactly the same way and be the same size; they have to be congruent.0283

But here, because this image and this image are different sizes, but the same shape, this has to be dilation.0289

Think of dilation as...when something dilates, it gets bigger; so it is getting bigger or getting smaller.0309

The next example: Determine if the transformation is an isometry.0318

Remember: an isometry is when you have two images, and the pre-image and the image are congruent; that is for rotation, reflection, and translation.0322

We just want to see if these two are congruent.0335

Now, to determine if two triangles are congruent, remember: we have those theorems and postulate,0340

where it says Angle-Side-Angle (they are corresponding parts), Side-Angle-Side, Side-Angle-Angle, and Side-Side-Side.0346

Those are the different congruence theorems and postulate.0356

We want to see if this pair of triangles applies to any of those.0364

Now, here I see that an angle is congruent here, and a side, and they are corresponding parts.0371

Now, for this one, angle B is corresponding with angle E; this one is given, and this one is not.0381

And angle C is corresponding with angle F, but this one is not given, and this one is.0390

I want to find the measure of this angle, and I can do that by taking these two and subtracting it from 180; so it is 180 - (105 + 40).0396

This right here is 145; so if you subtract this from 180, you will get 35 degrees.0418

The measure of angle C is 35 degrees.0432

Now, the measure of angle E is going to be 180 - (35 + 40); now, we don't have to solve for that, because we know that this is 35;0439

this angle is congruent to this angle; and of course, that means that this angle has to be congruent to this angle.0458

So, I have Angle-Side-Angle, because this pair of angles is congruent; their sides are congruent; and the angles are congruent.0467

So, because of this, these two triangles are congruent, and therefore, this is an isometry; so it is "yes."0481

The next one: Show that triangle ABC and triangle DEF are an isometry (so it is the same type of problem).0497

Now, for this, we have the coordinates of each vertex for each triangle.0508

So, we can find the measure, or the length, of each side.0520

I can just find the measure of that side with the length of that side and compare them and see if they are congruent.0528

I have to use the distance formula: the distance formula is (x2 - x1)2 + (y2 - y1)2.0536

Remember: this means the second x; so it is the second x, minus the first x, and the second y, minus the first y.0550

A is (-6,1); B is (-4,6); and C is (-2,3); then, D is (1,-1), E is (3,4); and F is (5,1); find the distance of AB.0559

So, AB is, let's see, (-4 + 6)2, and then (6 - 1)2.0608

I have that this is 2 squared, plus 5 squared, which is 4 + 25, which is √29.0626

And then, let's do DE: DE is (3 - 1)2 + (4 + 1)2; and it is plus because it is 4 - -1.0649

So, this is 2 squared, plus 5 squared, the same as that; so it is the square root of 29.0668

I know that AB is congruent to DE; now, BC is (-2 + 4)2 + (3 - 6)2;0678

so, this is 2 squared, plus -3 squared, which is 4 +...this is 9; that is √13.0704

And then, what is corresponding with BC? EF.0721

EF is (5 - 3)2 + (1 - 4)2: 2 squared plus -3 squared is the same, √13; so those two are the same.0725

And then, AC (I am running out of room here) is (-2 + 6)2 + (3 - 1)2.0753

That is 42 + 22; this is 16 + 4, is 20; √20...0781

And then, AC and DF...DF is (5 - 1)2 + (1 + 1)2, so this is 42 + 22,0794

16 + 4; that is 20, so that is √20; so then, these two are the same.0827

So, by the Side-Side-Side Congruence Theorem, they are congruent, which means that it is an isometry.0838

That is it for this lesson; thank you for watching Educator.com.0850

Welcome back to Educator.com.0000

For the next lesson, we are going to go over reflections.0002

Remember: a reflection is a type of transformation whose image and pre-image mirror each other.0007

It is a congruence transformation, meaning that, when you have the pre-image,0014

and you reflect it to create the new image, they are going to be congruent; those two images are going to be exactly the same.0020

And again, reflection is like a mirror; think of them as reflecting each other.0029

That line that acts as a mirror, the line that creates the reflection from the pre-image to the image, is the line of reflection.0038

This is like the mirror itself; here, if this is an image, and this is the pre-image, this is the line of reflection.0053

To draw the new image, you would have to draw exactly on the other side.0067

So, if this was the mirror, then it would reflect on the other side, the same exact way.0072

This would be called the line of reflection.0078

And the point of reflection is the point that reflects both images.0083

If this is the point of reflection, then it would have to reflect this image on the other side.0091

So, here it is the same distance; so if it goes that much that way from the pre-image, then it has to go the same distance to the other side to create the image.0099

Here, this is the line of reflection, and this is the point of reflection.0114

Now, a symmetry: we have line symmetry and point symmetry; line symmetry is kind of like that line of reflection,0122

where it creates two halves: it is the line that makes it symmetric for both sides.0134

This equilateral triangle has three lines of symmetry, because you can draw it here to create the two equal halves, this half and this half.0145

This is another one, and another; so all three would be lines of symmetry.0162

This one is the point of symmetry, because if you go this way, the distance to that point on the image0170

will be the same as if you go the opposite way to that point on the image.0186

That is point symmetry; and images have point symmetry, or they do not have point symmetry.0192

This one does, because no matter where I go...I can go here...then when I go the opposite way,0199

it is going to be exactly the same distance away from that point.0206

So, the same thing happens to those vertices; I can go this way there, and then this way there--it is exactly the same; that is point symmetry.0210

So here, we are going to draw the image over the line of reflection and the point of reflection.0227

Here is the line of reflection; here is the point of reflection; I want to reflect this image on this line and on this point.0232

To do this, it is best to use their vertices, to reflect the vertices instead of just trying to draw the image (it is not going to be accurate).0244

Start with the points: here, if I reflect this point along this line, then it is going to be around here somewhere, around there.0256

So, this will be...whenever you create a new image, you are going to call it a "prime"; it looks like an apostrophe: A': it is the same, but it is the prime.0269

And then, for C, it is going to go maybe this much; let's call that C'.0282

And then, for B, I am going to go maybe this much; so this will be B'.0294

And then, just connect them; make your sides; that would be a pre-image, and then this is the image.0306

So now, to reflect this on this point of reflection, you are not going straight across, like you did for this line.0323

For this one, you are going to go from B; you are not going to go this way; instead, if you want, you can use a ruler;0335

and you are going to go directly in the direction of that point, and then you go that same distance from the other side of that point, like that.0343

That would be B'; and the same thing happens here, for C; you are going to go directly towards that point of reflection,0355

and then keep going to the other side that same distance; there is C';0370

and then, for A, go about that much, so it would be right there; so here is A'.0380

Then, draw your triangle; there is the reflection along the point of reflection; so here is the first one, and here is the second one.0396

Determine how many lines of symmetry each figure has, and identify whether each figure has point symmetry.0422

This first one is a circle; now, where can I draw a line that will create symmetry for this image?0429

I can draw a line here; I can draw a line here; I can draw a line here.0439

A circle has infinite lines of symmetry, because I can draw a line through this circle anywhere I want,0450

as long as it is passing through the center; and then, each of those will be a line of symmetry.0480

This one, I know, I can cut down this way; and I can cut it down that way, in half--I will have two equal parts.0489

Can I cut it this way, diagonally?--no, because if I cut it diagonally, even though it is going to be two equal halves,0503

it is not going to be symmetric; it is not going to be exactly the same on both sides;0510

only these two work, so this one has two lines of symmetry.0515

This one...let's see; can we draw it this way?--no, because these are not the same; they are not the same length.0529

It looks like I wouldn't be able to cut it anywhere to make it symmetric, so this has no lines of symmetry--none.0542

Now, for point symmetry (I am going to use red for that), can we draw a point within this image so that,0559

no matter which way I go, both sides of the point are going to be exactly the same distance to the image, like that and then like that?0571

Or like this...is that going to be the same distance as that?0585

And this is "yes"; this has point symmetry.0591

How about this one--if I draw a point right there in the center, if I go this way, is that the same distance as if I go this way?0597

How about if I go there--is that the same distance as from this diagonal?0610

It looks like it, so I am going to go here and go here; this one is "yes."0618

And then, for this one, let's say I am trying to find as close to the center as possible--somewhere right here.0626

If I go this way, is that going to be the same distance as if I go this way?0638

No, this is the center; we can tell that this is a lot longer, a lot further away, than to this point; this is "no point symmetry."0648

If I go this way, that is not the same distance as from the center to this point; so this is "no"--this does not have point symmetry.0662

We are going to graph some reflections: Graph the reflection of the polygon in the line y = x.0679

For this line, we know...we will use red to graph the line of reflection...that the y-intercept is 0; we will plot 0.0686

The slope is 1, so that means that those are the point of my line of reflection, so I am going to graph this line.0699

We are going to reflect it along that line; so again, to reflect this, do not just reflect the image.0720

You want to reflect each of the points, and then you can graph your image.0729

The first point: let's do D; D is right here; so then, we can go right here for the D.0735

If you are going to draw lines to guide you to find the distance away this way and the distance away that way,0752

then make sure it is perpendicular to that line of reflection.0758

This is D'; for C, see how I go diagonally that way; it is negative slope; right there, this is C'.0765

For this one, this line that I am going draw has a negative slope, so it is as if it is going diagonally 1, 2, 3, 4, and a half;0785

here is the half; 1, 2, 3, 4, right here...this is B'.0806

And then, for A, it is going to go 1, 2, 3, 1, 2, 3; here is A'; so, my image is right here.0815

OK, now, to find the coordinates of my pre-image, this one in black, I know that A is...0838

let me do it down here...(-4,2); B is (-4,5); C is (1,5); and D is (1,2).0850

Now, for the reflected image, A' became (2,-4); B' became (5,-4); and C' is (5,1); D' is (2,1).0879

OK, do you notice something about these coordinates and these coordinates?0909

Here, this is (x,y); well, the x in the pre-image became the y in the new image, so it is as if we just flipped it: (y,x).0916

When you are reflecting along this line, y = x, you just flip x and y.0938

Whenever you reflect along y = x or maybe the y-axis or the x-axis, which we are going to do in the next example,0948

it is always going to be something with the coordinates.0958

This one, we are going to reflect; we are going to draw; it is going to be a triangle; and then, we are going to reflect over the x-axis and then the y-axis.0969

We are going to reflect it twice.0978

In the pre-image, A, is (2,0); B is (4,2); so here is A, and here is B; and then, C is (3,-1).0984

Now, we are going to reflect over the x-axis; so if this is the mirror, we are going to reflect along this.1010

Now, when we reflect along this line right here, this point is on that line.1021

This point is on the line of reflection; if it is on the line of reflection, then it doesn't move anywhere; it stays there.1029

This is C; so then, this is going to stay there as A'; then, point C is going to reflect to right there, so this is C';1036

and then, B is going to reflect two down, so that is going to be 2 this way; this is B'.1052

It goes A', B', C'; there is our reflected image, when we reflect along the x-axis.1062

I know that it looks kind of confusing, but here, again, if a point is on the line of reflection, then it stays there;1078

then the pre-image and the image point is going to stay on that line.1089

And then, this one was on the other side; see how this one is on this side and this one is on the other side.1094

Well, it just has to go to the opposite side of the line of reflection; so this is that new image.1099

And then, we are going to reflect this image along the y-axis, and that is going to be the blue image;1108

and then, I am going to draw this new image in red.1118

This is the line of reflection, the y-axis; so A' is now going to go here, again, because this is acting as the mirror.1124

And this is now called "A double prime," because I reflected it for the second time.1136

And then here, 1, 2, 3, 4, 1, 2, 3, 4...here, this is my B double prime, and C...away from this line is 3 spaces, 3 units, so it is 1, 2, 3 units right here: C''.1145

It is going to go like that: the black was my pre-image, and the blue was the prime--the first new image.1171

And then, this became double-prime; that is the second image.1191

It reflected on the x-axis, and then it reflects along the y-axis.1195

OK, now, if you were to draw that, that is what happens.1201

Let's look at the coordinates now: A is (2,0); B is (4,2); and C is (3,-1).1207

For the blue image, A' became (2,0); B' is (4,-2); and C' is (3,1).1229

Reflecting on the y-axis, this was a reflection of the blue; the red, the reflection along the y-axis, was a reflection of this one right here.1255

A'' is (-2,0); B'' is (-4,-2); and then, C'' is (-3,1).1275

If you notice this one right here, see how the y became negative (see the difference right here).1295

So, when you reflect along the x-axis, then the y becomes -y.1314

This is 0, so it will stay the same; and then, 2 became -2, and -1 became 1.1327

When you reflect along the y-axis, then notice how the x changed; the y stayed the same, and the x-coordinates became negative.1333

Well, those are good if you want to remember it that way.1355

So, if you were given coordinates, ordered pairs to reflect and find the new coordinates,1361

if you reflect along the x-axis, if it says to reflect along the x-axis, keep the coordinates the same; just make the y negative.1370

And if it says to reflect along the y-axis, the new coordinates would just be the x-coordinates becoming negative.1377

If you ever have to reflect along the line y = x, then you have to switch x and y.1387

Well, that is it for this lesson; thank you for watching Educator.com.1394

Welcome back to Educator.com.0000

For the next lesson, we are going to go over translations.0002

Remember: a translation is a congruence transformation where all of the points of an image are moved from one place to another place without changing.0006

So, it is just the same distance and the same direction--the same image; it just moves.0016

Remember: another word for translations would be "sliding"; slide, shift, glide--those are all words for translation.0026

The first image, the initial image, is called the pre-image, and then it goes to the new image.0039

I like to say "new image," because that is the image created from the pre-image.0050

If the pre-image has coordinates (x,y) for point A, then the new image is going to be A',0058

and it is the same (x,y) coordinates, but shifted up or down, and that would be a...0069

I'm sorry; a would be left and right, because x is moving horizontally.0078

So then, a would be how many it is moving either left or right; and b,0085

because it is the y-coordinate plus the b, is how many the y-coordinate is moving up and down.0090

So then, this would be A'; so let's say this is the pre-image, and we have A (let's say this is A), and the coordinate for this are (2,1).0098

Now, if this whole image, the pre-image, shifted up 1, or let's just say it shifted right 4 and up 1;0111

we know that, if it shifts right, it is a positive 4; if it shifts up, then it is a positive 1.0126

If it shifts to the left, then we know that it is a negative; so going this way, left is negative, and down is negative; these are negative numbers.0132

These are positive numbers; so if A is shifting to the right 4 and up 1, then I can say that the x-coordinate0144

(this has to do with the x, if it is moving left and right)...for A, it is (x,y); A' is x + a...lowercase,0155

just so that you know that it is not the same as this coordinate, A, and then y + b.0170

How many did it move left or right? It moved 4 to the right, so this point, (2,1), became...0177

for A', x is 2; we went 4, and then y moved up 1, so that is y, which is 1, plus 1; so then, my A' is (6,2).0188

All it is: (2,1) became 2 + 4...this is how much it moved...and 1 + 1; so it is the same coordinates,0222

but then we just count how many we shift: 4 to the right and 1 up--so our new coordinates are going to be right here, so it will be A'.0231

(6,2) will be the new coordinates for A.0250

If we do another one, let's say that this A is (-2,-1); let's say it shifted; it moved left 2 and down 3.0257

So then, my A', we know, is -2 + a, and -1 + b; see how it is the same coordinates here, -1 and 2, and we just add how many we move to this coordinate.0296

How many did it move left and right? It moved left 2.0325

Again, if we move this way on the coordinate plane, then it is a negative number, because we are moving towards a negative.0328

So then, it is going to be -2 -2, because for a, how many did I move?...-2, and then -1 - 3, because I moved down 3.0336

If I go down, this is going towards the negative numbers; this is positive; this is negative; this is negative; and this is positive.0353

A' is (-4,-4); if you take this ordered pair, and we move 2 to the left and down 3, then I will be at (-4,-4).0367

Now, translations are actually a composite of two reflections.0390

If this is the pre-image, and you reflect it once, you get this image; this is reflection, like the mirror reflection.0399

When we reflect it again (and again, the lines of reflection have to be parallel), reflect it a second time,0412

then this image right here, this pre-image, to this image right here, is actually a translation.0420

Two reflections is equal to one translation, but of course, only when the two lines are parallel.0429

If the two lines that you reflected along are not parallel, then it is not going to be a translation.0446

Only if the two lines are parallel when you make your two reflections, then it becomes a translation.0455

That is a composite of reflections, or a composition of reflections.0463

We are going to use this translation for each ordered pair; that is the rule.0470

Now, just to show you, if you are still confused about this, A is (4,0), so 1, 2, 3, 4...0; this is where A is at.0481

Now, what this is saying is that you are going to take the x-coordinate, and we are going to add 3.0497

We are going to move 3; now, it is the x number, so if it is the x number, and we are going to move 3, then it has to go left or right,0503

because x's are only along this line; you can't count up and down and call that x.0515

It is the x-coordinate, which is the 4, and we are going to add 3 to that.0523

So, if you add 4, that will be x, which is 4, plus 3; and then, the y, which is 0, minus 4; then A' becomes (7,-4).0528

And again, it is just moving; that means that it is saying, "Well, we are going to add 3; so we are going to go 1, 2, 3 to the right,0552

because that is a positive"; this becomes A', and then you are going to go down 1, 2, 3, 4, down the y-coordinate;0559

it was at 0; now it is going to go to -4; so go 3 to the right and 4 down.0581

This is the left-right number; this is the up-down number; see how many you are going to move--that is all that it is saying.0590

There are your new coordinates for the prime.0597

Now, for this one, we are going backwards, because we have the prime number, so we want to just find B.0608

That means that, for this B, we have (x,y), and then B' became (x + a, y + b).0615

And we know that B' is (-5,-2); so, I want to find out what my (x,y) is, because if I have my (x,y), then that will give me B.0637

How did I get -5? That is x + a; so the x-coordinate, including how many we move left and right, is going to become -5.0656

So, I know that x + a equals -5; isn't this equal to this?0666

And then, the same thing happens here: y + b equals -2.0678

Now, I know my a and my b; my a is positive 3, so x + 3 equals -5; if I subtract 3, then I get that x is -8.0685

Here, this is y + -4 = -2; add the 4, so y is 2.0702

My B, then, becomes (-8,2); so again, I just took this rule: it is (x + 3, y - 4); and you just make this x + 3 equal to -5,0718

which is right here; and then, make y - 4 equal to -2, because this is our B', and then this is our B.0740

You want to use this to solve for that, to find the x and the y.0752

OK, given the coordinates of the image and the pre-image, find the rule for the translation.0760

This is the pre-image, and this is the new image; find the rule for the translation.0767

I want to know how to get from R (we are going to call that (x,y); that is the pre-image),0774

and then it went to R', when you add a to the x and you add b to the y.0783

R is (-2,8), and it became R', which was...let me just replace this (x,y) with this (x,y) in here: it was -2 + a, and then 8 + b.0796

The R', we know, is 4 and -4; so if -2 + a is the x-coordinate for R', and that is 4, then isn't that the same thing as that?0822

I can say that -2 + a is equal to 4; and the same thing works here: 8 + b is equal to -4.0839

So here, I am going to add 2; a is equal to 6; and then here, if I subtract the 8, b is equal to -12.0850

They want to know the rule for the translation: the rule would be back up here; this is what we used for the rule.0866

We are going to keep everything the same; we are just going to replace a and b.0875

The rule for R (we have coordinates (x,y)), to make it into R', is x + 6, comma, y - 12.0879

So, whatever this point is, we are going to move 6 to the right, and we are going to move 12 units down; that is the rule.0894

This would be the final answer.0902

OK, and the next one: Find the translation image using a composite of reflections.0910

Remember: we need two reflections along parallel lines to make those two reflections equal to one translation.0916

Here, we have two parallel lines; I want to reflect this image twice--the first one along this line and the second one along this line.0926

And then, I will get one translation, meaning that, for my second reflection, it should look exactly like this.0935

So the, the first reflection: remember: to draw reflections, you are just going to reflect the points.0945

That is this point right here; for this one, you are going to go out this much; and for this point, you are going to go right there.0951

It is going to be the first reflection, and then the second reflection, like this; and then, that is maybe right there.0966

And then, the same thing happens for this; it is going to go (I'll draw the line better) here.0986

See, if you only look at this black and this red image, it is a translation.1003

All it did: it is as if this just glided over to this side; it is the same; it didn't rotate; it didn't flip; it didn't do anything else but just slide over there.1014

Again, two reflections equals one translation, as long as the two lines are parallel.1025

If they are not parallel, then it is not going to be a translation.1033

In the fourth example, find the value of each variable in the translation.1040

Here, we have that one of these is the pre-image and one of them is the translated image.1046

And because it is a congruence transformation, they have to be exactly the same.1053

They are congruent, meaning that all corresponding parts are congruent.1059

We just want to find the value of each variable; we have x here, and we have y here.1063

That means that this angle and that angle are corresponding, and they are congruent, so I can make them equal to each other.1071

So, 2x is equal to 80; to solve for x, I just divide the 2; so x is 40.1081

And then, for my y, to make this angle and that angle congruent, y is equal to 110.1096

And again, the whole point of this is to just keep in mind that translation is a congruence transformation.1107

So, always remember that all corresponding parts are congruent.1115

That is it for this lesson; thank you for watching Educator.com.1121

Welcome back to Educator.com.0000

For the next lesson, we are going to go over rotations.0002

Now, rotations, remember, are a type of congruence transformation.0006

This is the third one, in which an image moves in a circular motion to a new position.0011

Here, this is our pre-image; the center of rotation is a point (that is this right here--this is the center of rotation);0017

it is a fixed point which the image is rotating around.0034

Remember: rotation is when we are rotating it; we are going either this way, clockwise, or counterclockwise, to create a congruent image.0037

This image that we start with is always called the pre-image; this is the fixed point that is the center of rotation;0052

this is the point that we are going to rotate around.0058

The way we do that, and the easiest way to do rotations, is to pick a point; pick one of the vertices here,0063

on the pre-image (I am going to pick that point right there), and you are going to draw a line to the center of rotation.0076

Now, the center of rotation, when we draw a line like this, is going to create an angle.0087

So, if we are going to draw a line from this fixed point to the center of rotation, and then draw another line0094

to the corresponding point in the new image, which is this right here--this is corresponding to this point right here--0101

that is going to have an angle; so this is the angle of rotation.0111

You are moving a certain angle amount, a certain degree, either clockwise or counterclockwise, from the pre-image to the image.0116

Remember: this is a congruence transformation, so both images are congruent; they are the same.0125

There are two ways in which you can perform a rotation.0135

The first is a composite of two successive reflections over two intersecting lines.0140

Remember: for translations, we also performed two reflections over two parallel lines; don't get that confused.0151

For translation, it is also two reflections, but with two parallel lines; here, for rotation, it is also two reflections,0165

but the lines are intersecting; they have to be intersecting.0175

If this is our pre-image, it is first reflected over this line right here to get this.0182

So, if I call this, let's say, A, then this corresponding point, which is right here (because it is a reflection;0197

remember that a reflection is like the mirror, where you are making this act as a mirror,0206

and you are reflecting this image)--if this is A, then this would be A'.0211

This is the first reflection, and then you are going to do a second reflection over this line.0218

This point now becomes this point way over here; so this is A''; that means that this pre-image to this image right here, A'', is a rotation.0224

We rotated this image to this image right here.0242

So again, to do a rotation, one rotation can be the same as two reflections, but with intersecting lines.0247

The lines of reflection have to be intersecting; that is the first way.0265

The second way to perform a rotation is kind of how we did on the previous slide, where you are going to have an angle;0271

so then, you are going to pick a point from the pre-image, draw a line to that center of rotation,0282

and then find the corresponding point in the other image, right there; and then, you are going to draw a line to that point.0290

And this angle that is formed is called the angle of rotation.0299

Whenever you do angle of rotation, I will always do it in red, so that you know that that is what we are doing.0307

We are finding the angle from the corresponding point to the center of rotation, and then back to the image that was formed.0312

That is the angle of rotation; and then, this is the same diagram from the previous slide, and this is where we get our postulate.0323

In a given rotation, the angle of rotation is twice the measure of the angle formed by the intersecting lines of reflection.0341

This shows you both methods in one diagram: remember: with the first method, we did two reflections;0350

this one reflected to this; this reflected to this; and this became A''.0359

And then, the second method: we did, right here, the angle of rotation.0364

This shows you both methods in one diagram, and it is saying that this angle right here,0378

the angle between the two intersecting lines of reflection--if this is 90 (let's say that it is a right angle),0386

then the angle of rotation is going to be double that; so the angle between the lines of reflection,0396

the angle formed by method 1, is going to be half the measure of the angle formed from the angle of rotation,0408

this whole thing right here, which was method 2.0416

So then again, the angle of rotation, this right here, the red, is going to be double the blue, between the lines of reflection.0424

The angle of rotation is twice the measure of the angle formed from the lines of reflection.0433

Then, this angle of rotation is going to be 180; and that is the postulate on rotation.0442

Then, our first example: Find the rotated image by reflecting the image twice over the intersecting lines.0455

Remember: one rotation is going to be two reflections only if the lines of reflection are intersecting.0462

So then, here is our pre-image, and we are going to reflect it along here, and then, for the second reflection, reflect it along this one.0479

So, we are going to get our new image here.0488

Remember: to reflect, you are going to go on the other side like that, here, something like that right there.0492

And then, for this one, put it maybe right there; it is going to be that.0508

If this is A, then this is A'; that is the first reflection.0529

Then, for the second reflection, we are going to reflect over this one right here; it is going to be right there.0536

And then, let me just do that one in red, so that you know that this is the second reflection...right there.0546

And then, for this one, make sure that, when you draw your line of reflection, it is perpendicular.0557

So then, moving straight across...maybe it is somewhere right there...it is going to be something like that.0570

This is corresponding to that point right there; that is A''.0585

Make sure that, if you are going to draw these lines to help you, to guide you in where to draw, that they are kind of faint.0594

You can either draw them dotted, or maybe you can erase them after, because they are not really supposed to be there.0604

This is the image right here; this would be the reflected image.0611

So, from this pre-image, the rotation occurs to this image right here; from this, and then to this, is the rotation.0616

The next example: Triangle ABC is rotated 180 degrees on the coordinate plane; draw out a form of triangle A'B'C'.0634

So, we are going to rotate it 180 degrees.0645

Now, it doesn't tell us which way to go, clockwise or counterclockwise.0648

But for this problem, it doesn't matter, because 180 degrees is just a straight line.0653

So, whether you go this way or the other way, it doesn't matter; you are going to end up in the same spot.0659

If we say that the origin is the center of rotation, this is our fixed point; this is where we are going to rotate around.0668

Since we know that this is the angle of rotation, that is how much we are going to be moving.0680

Your fixed point (and again, I am doing this in red, so that you know that this is the angle of rotation, method #2):0691

you are going to draw a line to your fixed point, your center of rotation, and then you are going to go the same distance.0696

And make sure that this angle is the angle of rotation, which is 180.0708

So then, this is going to end up right there; so this is C'.0717

And then, for B, right there, it is going to go...you can also use slope to help you,0726

because if it is 180 degrees, then you have to make sure that it is a straight line;0739

so this angle of rotation has to be a straight line; so then, the slope of this right here is 1/4;0744

so then, I have to make sure that, when I go this way, when I keep going, it is also going to be 1/4.0752

This right here is B'; and again, you have to make sure that it is a straight line.0760

If you want to use a ruler, you can use a ruler, because if you go here, and then you can maybe go a little bit sideways,0768

then you can end up anywhere but there; so make sure that it is a straight line, 180 degrees.0776

That is B'; and then, for A, again, you are going to draw like that; my slope of this is 5/2.0783

So then, I have to make sure that my slope is also going to be 5/2; it is as if it is going this way in a straight line, 5/2.0797

This right here is A': this is C', this is B', and this is A'.0813

Usually, rotation is probably going to be the most difficult to draw, but instead of looking at this whole diagram,0836

instead of trying to draw that image all at once, use the points.0845

And then, all you are doing is plotting a new place for this point: the rotated image of this point,0853

the rotated image of that point, and the same thing for that point.0860

And then, you just connect them to create your new image.0863

Find the value of each variable in the rotation: in this one, here is the pre-image--I know that because it rotated this way 90 degrees.0871

It is a rotation, and for this, we are just finding the value; it is just to show that these are congruent, that rotation is a congruence transformation.0884

The first equation that I can set is with this right here, the x.0900

This is corresponding, on the pre-image, to that right there; so x equals y - 5.0906

I have two variables that I need to make my other equation; here is y; y is corresponding to this right here, so that is 15.0917

So, that is one of my variables; and then, I need to plug this into this equation right here, so that I can find x.0930

x equals...y is 15, minus 5; so x is 10.0936

If you get a problem like this, where you have to show that they are congruent, then just set your equations.0947

Make sure that you make this equal to this; it is not x with the 15, so be careful on what you make equal to each other.0958

It has to be the corresponding sides and angles.0969

Draw the polygon rotated 90 degrees clockwise about P; this is P right here.0976

And we are going to draw it clockwise, meaning that we are going to go this way, 90 degrees.0987

If you want to, you can use a protractor; we are going to just sketch it.0998

I am going to do one point at a time, one vertex at a time.1007

Here, point C--I am going to draw it like this, and then I am going to draw my other line to my image, to C',1013

so that the angle formed is going to be 90 degrees.1027

So then, maybe it is right there; so then, here, make sure that it is the same distance; that is C'.1032

Let's do B: here, it is something like that; and then, maybe it is right there somewhere; this is going to be B'.1050

Then, my A: this one is going to be a little bit further, so it is better for you to use a ruler, because it is kind of far.1080

Just use that as a reference; you can also use maybe the distance between this line and then to the C,1095

so then this line to this line--how far apart are they?--and then make it that same distance when we draw this line.1101

For this one, it is going to go somewhere like that.1109

And again, make sure that this line right here, to this line, AP to PA', is a 90-degree angle.1119

It would be somewhere like right there; it doesn't look straight--let me just draw that again.1137

I am sure that my screen is a lot bigger than your paper, so it should be easier for you.1148

I am going to label that A', and then for D...I just made that wrong; let's see,1162

it should actually be going somewhere in front of B, somewhere there: A'...1192

And then, for D, it should be somewhere there; so this is D', and then this is A'.1212

When you draw it, you should have the same figure as this; it should look congruent.1228

Again, if you have a ruler, that will be a lot easier; I didn't have a ruler, so it was a little bit harder for me.1242

Just make sure that when you draw the line from the point to the center of rotation,1252

and then from the center of rotation to that prime, that it is a right angle; it has to be 90 degrees,1258

or whatever the angle of rotation is; if it is 100 degrees, then make sure that it is about 100 degrees.1264

And then, your pre-image and your image have to look the same.1272

This is just a sketch, so it is not exactly the same; but it should look very close to it.1276

Well, that is it for this lesson; thank you for watching Educator.com.1284

Welcome back to Educator.com.0000

For the next lesson, we are going to go over the fourth and final transformation, and that is dilation.0002

Dilation is a transformation that alters the size of a geometric figure, but does not change the shape.0009

The first three transformations that we went over (those were translation, reflection, and rotation) are all congruence transformations,0017

meaning that when you perform that transformation, the pre-image and the image are exactly the same; they are congruent.0026

Dilation is the only transformation that is not a congruence transformation.0034

Now, it says that it alters the size, but not the shape; so that means that the shape is the same, but the size can change.0041

That, we know, is similarity; so for dilation, the pre-image and the image are going to be similar.0052

They are not going to be congruent; they could be, if they have the same ratio;0062

but otherwise, the pre-image and the image are going to be similar.0068

If this is the pre-image and this is the dilated image, then it got smaller from the pre-image to the image.0075

That is what is called a reduction: it got smaller.0084

If this is the pre-image, and this is the dilated image, then it got bigger, so it is an enlargement.0088

So, we are going to go over that next, which is scale factor.0095

Now, the scale factor (we are going to use k as the scale factor) is the ratio between the image and the pre-image.0100

The image is the dilated image; it is the new image; the pre-image is the original.0119

So, any time you see "prime"--here we see A'--that has to do with the new image, the dilated image.0126

And then, we know that this is the pre-image.0138

For a dilation, we are going to have a center; this is the center, C.0141

And we are going to base our dilation (meaning our enlargement/reduction) on this center.0148

Now, again, the scale factor is the image to the pre-image.0156

We can also think of it as from the center to...and then, which one is the image, this one or this one?0169

We know that it is the prime; whenever you see the prime, that is the clear indication that it is going to be part of the image.0178

So, if I want to measure the length from C, the center, to the image, that point right there, that is going to be CA', that segment.0184

That has to do with the image; so that is going to be the numerator, over...C to the pre-image is that point right there, so CA.0204

So, it is going to be CA' to CA.0216

Now, if I were to draw a line from C to A' to show the length, this center, the image, and the pre-image are all going to line up.0222

It is always going to line up; that is what we are going to base it on.0246

So, we are going to use that to draw some dilations: so again, it is the center to the image0249

--anything that says "prime"--that length, over the center to the pre-image.0261

This, we know, is an enlargement, because this was the pre-image, and this is the dilated image; it got bigger.0271

So, from here to here, it is bigger; so my scale factor for this one...if I say that k is 2, then it is actually 2/1.0279

So then, this number up here has to do with my image, and this number down here has to do with my pre-image.0296

That means that my image is twice as big, or as long, as my pre-image; that is scale factor--that is what it is talking about.0302

It is comparing the image to the pre-image; so I can also use this ratio for the length, the distance, from the center.0314

If all of this, from here to here, is 2, then from my center, the distance away from the center of the pre-image is going to be 1.0330

That is also talking about the scale factor; not only is it talking about the length or the size of the image and the pre-image,0346

but it is talking about the distance away from the center.0356

So, if this distance, from here in the image to the center, is 2; then from the pre-image to the center is 1.0359

And they are always going to line up; so the center, to B, to B'...they are all going to line up.0368

This one is an enlargement: it got bigger--it is twice as big as that.0377

Now, for the second diagram here, this is my center; this is A'; and this is A.0383

That means that this is my pre-image, and this is my dilated image.0392

Again, if I find the distance from my pre-image to that, it is all going to line up--the center, A', and A will line up.0404

Now, here, because the image right here, P to A'...let's say this is 1, and this is 1; that means that0422

from my center to my image is 1; and then, from the center to the pre-image has to be 2,0440

even though this part from here to here is 1; remember: it is from the center, so this point, all the way to the pre-image, is 2.0458

My scale factor is 1/2; now, if you notice, this is the pre-image (this is the original), and then this is my dilated image.0466

See how it got smaller: it is like saying, "Well, the image is half the size of the pre-image."0473

This is half the size of my pre-image; that is what the scale factor is saying--it is comparing these two.0483

And so, keep that in mind: this is image over pre-image, I/P.0491

Now, going over scale factor some more: if the scale factor is positive, then it is on the same side of the center point.0501

The two diagrams that we just went over were both on the same side, meaning that they were on the right side of the center.0514

Here is the center; this is to the right; and this is to the right...because the scale factor could be negative.0523

When it is positive, they are just on the same side; so then, the center to right here (let me always do that in red)...they are on the same side.0531

They are kind of going in the same direction; it is CA', over CA--it is always starting out from C, and it is going to there, and then C to A.0550

Now, if it is negative, then it is going to go to the opposite side of the center point; they are going to be on opposite sides.0568

So, here is A'--here is my image--and here is the pre-image.0575

Now, from the center point, C, if I go to the pre-image, it is going this way.0587

Now, let's say that my scale factor is -2: k = -2; so it is -2/1.0599

Because my scale factor is negative, I know that this is CA' over CA.0609

This right here, that I just drew, is this right here; so that means that this length right here is 1, because that is what that shows me: CA is 1.0622

Then, because CA' is negative, instead of going this way and then drawing it twice as big--0639

instead of going this way, I have to go the opposite way--that is what it is saying.0651

So then, if CA is going one way, that is your pre-image that is going this way, kind of to the left.0655

Then, this has to be drawn twice as long; so if CA is 1, then I have to draw CA' with a length of 2, but going in the opposite direction.0666

Just think of that negative as opposite; so then, if it went this way, then CA' is going to go twice as long.0677

From here to here is going to be 2; that is what it means to be negative.0687

So, if you have to draw your dilated image, that means that you are not going to have this.0696

You are just going to base it on this and this point; you are going in this direction--that is the pre-image.0702

So, when you draw your dilated image, instead of continuing on like you would here (this was your center, to the pre-image,0710

and then to draw your image, you kept going and had a line of C, A, A'--you are going to keep going0719

in that same direction if it is positive), because it is not positive, instead of going in the same direction,0725

we are going to turn around and go in the opposite direction.0730

And you are still going to draw it so that CA' is 2; so this is still 2--CA', that length right here, is 2.0734

It is the same thing, but it is going in the opposite direction.0745

And notice how C, A, and A' still line up, no matter what; if the scale factor is positive or negative, they are still going to all line up.0750

C, A, and A', C, A, and A'--they are all going to line up.0760

And then, next, we have enlargement and reduction--we kind of talked about this already.0766

If the absolute value of k (meaning regardless of if it is positive or negative) is bigger than 1, then it is going to be an enlargement,0773

because it is talking about the image and the pre-image.0784

So, if k, let's say, is 3, isn't it 3/1?0788

Isn't that saying that the image is 3 times bigger than the pre-image?--because we know that it is the image over the pre-image, I/P.0792

So then, if the image's length is 3, then the pre-image is 1, so then it has to be bigger,0802

because the number with the image is bigger than the number with the pre-image, so it is getting bigger--it is enlarging.0811

And that is just what this is saying; this is the pre-image, and this is the dilated image; it is getting bigger--small to bigger.0823

And then here, if the absolute value of k (meaning with no regard to whether it is positive or negative)...0833

if you have a fraction between 0 and 1 (let's say 1/2 or 1/3, or whatever...any fraction that is smaller than 1,0842

and greater than 0--it is going to be greater than 0, because it is absolute value), then it is going to be reduction,0853

because then you are saying that, let's say, for example, if k is 1/2, then again, this is image;0861

this number is associated with the image, and then this number is with the pre-image.0869

You are saying that the pre-image number is bigger than the image number.0877

If the image is 1, then the pre-image will be double that; so then, the pre-image,0882

the image before the dilation, is bigger than the dilated image; it is actually getting smaller--that is called reduction.0886

From C (if we are going to say that this is C), this is the pre-image; it is still going to line up.0896

That means that we know that A' has to be on that line; but it can't go this way, so it is going to be halfway between.0910

That means that, because, again, image is going to be CA'/CA, then this number right here...0919

if that is 1, then it is going to be 1 over whatever this whole thing is here, CA.0931

A review: If your scale factor, k, is positive, then you are going to keep drawing it in the same direction from the center.0939

So, it is on the same side of the center.0949

If it is negative, well, C to the pre-image is going in one direction; then, to go to the dilated image, you are going to go in the opposite direction.0954

It is like you are going to turn around if it is negative.0964

And then, regardless of it is positive or negative, if the absolute value is greater than 1, then it is going to be an enlargement,0969

because that means that the dilated image is larger than the pre-image.0978

And then, if it is between 0 and 1, then the top number is going to be smaller than the bottom number.0986

That means that the image is going to be smaller than the pre-image.0993

Let's do our examples: Find the scale factor used for the dilation with center C and determine if it is an enlargement or a reduction.1002

Here are our two similar figures, STUV and...here is the other image, because this has T' and U'.1015

So, I know that this bigger one is the pre-image; remember: it is always image to pre-image.1028

We see that this has "prime," T', and that has to do with the image, the dilated image, the new image.1044

That is going to be this right here.1054

That means that we went from pre-image, which is STUV, to this prime.1057

Because it got smaller, we know that it is a reduction; for #1, it is a reduction.1068

To find my scale factor, I want to find the ratio (because it is proportional, because these are similar;1081

dilation is always similar): so, do I have corresponding parts?1095

I have this right here, with this right here; so I do have the lengths of corresponding sides.1103

This one has to do with my new image, my dilated image, 4; and then, this right here is my pre-image; that is 9.1112

It is going to be proportional; so, my image length is 4; in the pre-image, the corresponding side is 9.1122

Now, even though this also has to do with the length of my pre-image, I can't use that,1136

because I don't have the other corresponding side; I don't have the measure of that side right there, which is corresponding.1142

So, I have to use the corresponding pair, 4 and 9.1148

And be careful: it is not 9/4, because the image number has to go on top.1153

This is the image; this is the pre-image; so it is image over pre-image, 4/9; so this is the scale factor.1158

The next example: If AB is 16, find the measure of the dilation image of AB with a scale factor of 3/2.1174

AB is a line segment; let's say that that is A, and that is B; and this has a measure of 16.1189

Find the measure of the dilation image of AB with a scale factor of 3/2.1203

Now, remember: our scale factor is image over pre-image, or CA' over CA.1209

We are going to use this as a reference for our scale factor; we know that it is 3/2.1224

Since the number that has to do with the image, the new image, is greater than this number down here,1231

which is the pre-image, I know that it is an enlargement--it got bigger--because the new image is bigger than the pre-image.1238

This is enlargement; that means that this pre-image is going to get bigger; my new image is going to be bigger than this.1246

Let's draw a center point: if that is my center, C, this right here, CA, is this number.1260

So, CA (I should do that in red) is what? 3/2--that is the scale factor;1279

so, my CA, the number associated with my pre-image, is 2; that means that CA is 2.1296

That means that my CA' is going to be 3.1303

Now, I know, because it went from C to A in this direction, and my scale factor is positive...1310

that means that I am going to keep going in that same direction to draw A'.1317

That means that CA' is going to be 3, so I can't draw it twice as long as this--I can't draw another 2--1322

because I have to make sure that from C to A' is going to be 3.1332

So, if this is 2, well, let me just break this up into units, then; if this is 1, then this is 2.1339

So then, 1, 2, and then another one right here...it is 1, 2, 3 in the same direction; and then, this will be A',1347

because again, it is not from here to here; it is from C to A'; C to A' is 3.1363

That means that if this is 1, then this whole thing is 3; and I just found that from my scale factor.1370

So, CA' is 3; CA is 2; make sure that C, A, and A' all line up.1378

And then, the same thing works here: this is CB' over CB; this is also 3/2.1386

We have this, and then we are going to keep going in that same direction, because it is a positive.1412

So, CB, we know, is 2; that means that CB'...when I draw my B', it has to be 3.1417

So, if this is 1, and this is 2, then this is a little bit more...and that is C...another one more...that makes this whole thing 3, and this is B'.1427

From here to here is going to be my dilated image.1449

And then, to find the measure of it...now, it didn't say to draw it, but then, just in case1458

you would have to draw it on your homework, or you have problems where you have to draw it,1463

just keep in mind that if it is a positive scale factor, make sure that C, A, and A' all line up;1468

and it is all going to go in the same direction; and then just do that for each of the points.1476

And then, if this is 16, remember: the image to the pre-image...this is the image to the pre-image, so the scale factor is 3/2.1484

That is the ratio; it equals...and then again, the ratio between these two is going to be the same.1497

So then, AB, my new image, is going to go on the top, and that is what I am looking for--this x.1504

That is x, over my pre-image (is 16); so this is a proportion--I can solve this by cross-multiplying:1512

2 times x equals 3 times 16, or I can just do this in my head: this is 2 times 8 equals 16,1523

so 3 times 8 is going to be 24--it is just the equivalent fraction (3/2 is the same thing as 24/16).1535

If you want to just cross-multiply, then it would be 2 times x, 2x, is equal to 3 times 16, which is 48.1546

And then, divide the 2; x = 24.1556

So then, right here, this has a measure of 24; so AB is 24.1564

The next one: Find the coordinates of the image with a scale factor of 2 and the origin as the center of the dilation.1580

Here is the center that we are basing that on; and our scale factor is 2, which means it is 2/1.1589

And I want to write image over pre-image; and then, you can write center...1601

Well, we already have C, so let's label that as P: PA'/PA.1611

So then, the scale factor is 2/1 (let me just write that here, too, so that you know that this is 2, and then this is 1).1622

PA, going in that direction, is 1; that means that I have to draw PA' as 2--it is going to go 1, 2.1637

And again, you are not starting from here and going 2 more; you are starting back at P and then going 2.1652

This right here is A'; and then, to PB...if that is 1, then to PB' is 1, and then 2; so this is B'.1658

And then, PD--that is this--is 1; then, PD' is 2; so then here, it is 1, and you go another--that is D'.1689

Make sure that they line up: P, D, and D'.1710

And then, go from P to C...like that; make sure that your lines are straight.1714

You can also use slope to help you: here, you know how we went down one, and then 1, 2, 3, 4: that is a -1/4 slope.1726

So then, I can go another 1, 2, 3...and then that would be right here; so it is going to keep going this way: this is C'.1737

Then, my new image is going to be from here, all the way down to here, to there, and there, and then there.1753

Make sure that your image has the same shape as your pre-image; it is just going to have a different size, but it is going to be the same shape.1769

That is my image; and then, I want to find the coordinates.1781

So then, A' is going to be (0,2); B' is going to be (4,4); C' is...this is 6, 7, 8, so (8,-2); and D' is (4,-4): those are my coordinates.1786

Now, if you had to find the coordinates without graphing--if you were just given the scale factor,1822

and you had to find the new coordinates for it--let's look at the original.1835

Let's look at the pre-image, just ABCD, the pre-image: the coordinates for the pre-image,1838

before we changed it, before we dilated it, were: (0,1); B was (2,2); C was...where is C?...right there: it is (4,-1); and D is (2,-2).1848

We went from the pre-image to image: notice how our scale factor is 2.1875

That means that our image is twice as big as our pre-image.1880

Look at this: your image, your coordinate points, are twice as big as your pre-image coordinates.1885

This one is (0,1), and this is (0,2); (2,2), (4,4); (4,-1), (8,-2); (2,-2), (4,-2).1896

It is like you just multiplied everything by 2, by the scale factor.1906

So then again, if you are given the coordinates of the pre-image, and you have a scale factor of 2,1910

that means that your image is going to be twice as big as your pre-image;1917

so then, you just have to multiply your pre-image by 2 to get your image; so then, those are your coordinates.1920

And the final example: Graph the polygon with the vertices A, B, C; use the origin as the center of dilation, and a scale factor of 1/2.1930

Let's copy these points: A is (1,-2); B is (6,-1); C is (4,-3); this was A, B, and C...so our polygon is a triangle.1942

And then, using the origin as the center of the dilation, and a scale factor of 1/2...again, the scale factor, k, is image over pre-image.1978

If this is a little confusing, you can always just, instead of "image," write "prime" or "new image" or something like that.1995

That way, you know what coordinates go with which one--image or pre-image.2003

This is our pre-image; our scale factor is 1/2.2011

I am going to use P for my center; what I can do is...for the image, it is PA', PB', PC', all for the image.2019

And then, the pre-image is just PA, to the original.2034

And our scale factor is 1/2: that means that our pre-image is twice the measure of our image--the pre-image is going to be bigger.2039

That means that, since the scale factor is 1/2 (which is smaller than 1), it is going to be a reduction.2050

The pre-image, the original, is larger than the new image, so the new image is smaller.2057

That means that our new image is going to be smaller than this.2063

Here, draw...again, from here, it is going to be like this; so then, PA (that is that) is 2.2072

That means that we are going to say that this whole thing is 2.2082

That means that PA' is going to be half that; it is going to be 1.2086

For PA', I am going to label it right there, halfway, because this whole thing is 2; then this has to be 1; that is going to be A'.2091

And then, for here to here, for C, our slope is 1, 2, 3 for 1, 2, 3, 4.2107

So then, here, we can just estimate where our halfway point is going to be, because this PC is 2; that means that PC' has to be 1.2122

So, if this whole thing is 2, then C' is going to be right there, halfway.2133

This is C', and then, for PB, it is going to go like that.2140

Our slope is down 1, over 6; so then, remember: our scale factor is going to be half that.2154

If this whole thing is 2, I have to find halfway; so if this is down 1, then it is only going to be down a half, because, remember, it is half of that.2160

So, go down 1/2; and then, going right was 6--we went down 1, right 6.2170

So then, instead of going all the way to 6, I have to go just halfway, which is 3; so it is going to be half, down half, and right 3.2176

There is my B'; so my new image is from here to here to here.2185

And then, let's see: all we had to do is just graph the polygon, and then use the origin as the center and a scale factor of 1/2.2199

Again, if the scale factor is smaller than 1, then you know that it is going to be a reduction; it is going to be smaller.2208

If it is greater than 1, then we know that it is going to be bigger than this pre-image.2216

That is it for this lesson; thank you for watching Educator.com.2223

Welcome back to Educator.com.0000

This next lesson is on inductive reasoning.0001

For inductive reasoning, we deal with what is called conjectures; a conjecture is an educated guess.0008

When you look at several different situations, or maybe previous experiences, to come up with a final conclusion, then that would be inductive reasoning.0017

When you have repeated observations, or you look at patterns, those things would be considered inductive reasoning.0036

Basically, you are just looking at past experiences--anything that will lead you to some sort of conclusion is inductive reasoning.0042

Looking at patterns: if I have 4, 8, 16, 32, and I need to use inductive reasoning0057

to find the next several terms in the sequence, well, I can just see how these numbers came about,0067

and then I can just apply the same rule to find the next few numbers.0073

So, for this, 4, 8, 16, 32...how did you get from 4 to 8, 8 to 16, and 16 to 32?0079

Well, it looks like this was multiplied by 2; that means I would have to multiply by 2 to get my next answer.0091

So, if I multiply this number by 2, then I am going to get 64; if I multiply this by 2, then I am going to get 128; and so on.0103

Now, for this next one, I have to see...well, here is a triangle, a square, triangle, triangle, square, square...what will be my next shape?0121

Well, it went from 1 triangle, 1 square, to 2 triangles, 2 squares; so my conjecture will be that it will be 3 triangles, and then 3 squares.0134

Again, we are looking at patterns, or looking at some kind of repeated behavior, to come up with a conclusion (or what will happen next).0155

Now, a few more problems: if we make a conjecture about this, if AB = CD and CD = EF, what can I conclude?0169

Well, if AB is equal to CD, if I draw AB, here is AB; and here is CD;0184

so, if I see that this and this are the same, CD = EF, so this equals this; isn't it true that, if this equals this and this equals this...0198

doesn't that mean that AB will equal EF?--so that will be my conjecture: AB = EF.0217

Make a conjecture, given points A, B, and C: let me just draw out a coordinate plane.0229

A is (-1,0); it is right here; B is (0,2); it is right there; C is (1,4), which is right there.0256

If I look at this, A, B, and C line up; so my conjecture would be that points A, B, and C are collinear, because they are on the same line.0272

Counter-examples: just because you come up with a conjecture, you come up with some kind of conclusion,0299

based on what you see in your observations, based on the patterns, and so on, doesn't mean that it is going to be true.0306

Just because something happens 3, 4, 5, or 6 times in a row doesn't mean that it is going to happen again the next time.0313

Conjectures are not always true; and to prove that it is not always true, you have to provide a counter-example.0322

And a counter-example is the opposite of what you are trying to prove.0331

If you are trying to prove that something is true--let's say you saw something a few times,0336

and so you conclude--you make a conjecture--that the next time, it is going to happen again;0343

you can't prove that it is going to happen again just by showing that it happened.0349

You cannot prove something just by giving an example, because it might not happen the following time.0354

You might not be able to find a counter-example in order to prove that that is not true.0364

A counter-example is the opposite of what you are trying to prove.0373

Let's say, for example, that the first five cars you see today are black; does that mean that all cars are black?0376

That would be a conjecture; the conjecture would be that, since you saw five cars that are black...0385

my conjecture would be that the next car that I see will be black; and that might not be true.0392

In order to prove that whatever you concluded, your conjecture, is not true, you are going to provide a counter-example.0403

A counter-example would be to show an example of it not being true.0413

Now, let's go over a few examples of this: the first one: Any three points will form a triangle.0418

If I have three points like this, I know that it is going to form a triangle.0427

Is this conjecture true? It could be true, but just because I gave an example of it being true does not make this conjecture true,0440

because I know that three points will not always form a triangle.0451

And so, what I can do to prove that this is not true--to prove that it is false: I can give an example of when this is not true.0458

And that would be a counter-example: so three points that do not form a triangle...there are three points; they don't form a triangle.0466

This is my counter-example: by giving an example of when this is not true, I am proving my conjecture false.0481

So, this conjecture...sometimes it could be true, but it is not always true.0493

By showing an example of when it is not true, a counter-example--that is when you are proving the conjecture false.0502

The square of any number is greater than the original number: well, that could be true, but it is not always true.0512

To show that it is not always true, I need to provide a counter-example.0519

Let's say I have some numbers: let's say 2--if I square it...I am saying the square of any number,0523

so if I take a number, and I square it, then it is going to be greater than this original number.0531

If I square this, then it is going to be 4; well, this is greater than this number, the original number.0536

What if I have 0? If I square it, what do I get? 0.0546

If I have, let's say, 1/2, and I square this, I get 1/4.0553

Well, is 1/4 greater than the original number, 1/2? No, 1/4 is smaller than 1/2.0564

So, for this example, this is true; but this one is not true; this is false, and this is false, for this being greater than the original number.0571

0 is not greater than 0, and 1/4 is not greater than 1/2.0582

Here is a counter-example, and here is a counter-example, because these two examples show that,0594

if you square the number, then the answer is not going to be greater than the original number.0602

Counter-example, counter-example: by showing the counter-examples, I am proving that this conjecture is false.0610

Find the pattern and the next two terms in the sequence: 15 to 12, 9 to 6...the pattern here that I see is subtracting 3.0623

So, from here to here, I subtract 3; subtract 3; subtract 3; if you subtract 3 again, then you get 3; you get 0, -3, and so on.0637

This one: 1 to 2, 2 to 6, 6 to 24, and so on--this one seems a little tricky, but you just have to look at it.0655

Here, if I...let's see...1 to 2: I can either add 1; I can multiply by 2; here I can add 4, or I can multiply by 3;0667

here I can add something bigger...for this one, it doesn't seem like that would be the pattern;0683

so here it is multiplied by 4; here, multiply by 5; so then, for the next number, I can multiply by 6.0691

If I multiply this by 6, then I will get...let's see: 120 times 6 is 0, 12, 720.0706

And the next one...you can just multiply by 7.0721

This one right here: I have a square; then I have this shape in there; and then I have another one.0728

The next pattern will be...so then, here is the next step; that is up to there.0736

And then, my next one will be to draw a square within that square, like that.0750

And the next one would be to do the same thing; and you are going to draw another square inside.0758

OK, so you are given a statement, and you need to come up with a conjecture.0769

The given statement is that angle 1 and angle 2 are adjacent.0776

"Adjacent" means that the angles share a side and a vertex.0781

So, if I draw angles 1 and 2, they are adjacent; then what can I conclude?0787

Well, I conclude that angles 1 and 2 (since I know that "adjacent" means that they are next to each other) share a side;0803

so, angles 1 and 2 share a side and a vertex, because they are adjacent.0817

The next one: the given statement: line m (here is line m, and this is a line) is an angle bisector of angle ABC.0842

Here is an angle, ABC; and line m is an angle bisector--"bisector" means that it cuts in half.0855

So, this line cuts this angle ABC in half; that means that these two parts right here are the same--they are congruent.0867

If I label this angle 1 and this angle 2, what can I conclude?0881

If this line is an angle bisector, then these two parts right here, angle 1 and angle 2, I can say are congruent.0887

Or I could say that the measure of angle 1 equals the measure of angle 2.0895

Or I can say that angle 1 is congruent to angle 2.0902

That would be my conjector, since line m is an angle bisector of angle ABC.0909

The next example: Decide if each conjecture is true or false; if true, then explain why; and if it is false, then we have to give a counter-example.0918

Given that WX = XY, the conjecture is that W, X, and Y are collinear points.0932

The conjecture is saying that if I have WX...here is point W, point X, and point Y...W, X, and Y are collinear.0947

WX = XY; the conjecture is that these are collinear; is that always true?0960

Can you think of an example of when it is not true?0968

Well, what if you have WX, XY...WX is equal to XY; this still applies here, but it doesn't prove that the conjecture is true.0971

This is a counter-example, because this is an example of when my conjecture is false.0997

This one I know...this conjecture is false.1005

The next one: Given that x is an integer, the conjecture is that -x is negative.1012

So, if x is an integer (some integers are 2, 0, and -2; so these are x), the conjecture is that -x is negative.1019

If I make this negative, -2 is -x--is it negative?--these would be -x.1033

-2: this is true; this is an example of the conjecture; what about 0?1045

If I make it negative x, then that will be -0, which is just 0; and that is not a negative, so this is false.1055

And then here, -2: if I make it -x, then it is -(-2); well, that is a positive 2.1067

So, does that show that -x is negative? No, because this is positive 2.1076

So, in this case, this one is false; if x is an integer (these numbers right here), and you make those integers negative,1083

then the answer is going to be negative: in this case, this works.1097

If you make this number negative, it is not a negative; if you take the negative of this number, it becomes positive.1103

So, -x is positive in this case; so this one is also false, and then here is my counter-example, right here.1112

This one and this one are both counter-examples, because they show that this is not true; it is false.1122

Well, that is it for this lesson; thank you for watching Educator.com--see you next time!1135

Welcome back to Educator.com.0000

This next lesson is on conditional statements.0002

If/then statements are called conditional statements, or conditionals.0006

When you have a statement in the form of if something, then something else, then that is considered a conditional statement.0014

If you have a statement "I use an umbrella when it rains," you can rewrite it as a conditional in if/then form.0025

So, "If it is raining, then I use an umbrella": that would be the conditional of the statement "I use an umbrella when it rains."0034

When do you use an umbrella? When it rains, right? So, "If it is raining, then I use an umbrella."0044

And that would be considered a conditional statement.0051

If...this part right here, "If it is raining"--the phrase after the "if" is called the hypothesis.0057

And then, the statement after the "then" is called the conclusion.0072

If it is raining, then I use an umbrella: this part right here is known as the hypothesis; "then I use an umbrella"--that is the conclusion.0080

That is what is going to result from the hypothesis.0087

You can also think of the hypothesis as p; p is the hypothesis, and q is the conclusion.0091

You can write this as a statement if p, then q, because p is the hypothesis; so it is if the hypothesis, then the conclusion.0105

And as symbols, you can write it like this: p → q; p implies q, and that would be the symbol for this condition, "if p, then q."0118

Again, the statement after the "if" is the hypothesis; the statement after the "then" is the conclusion.0134

And then, it is if p, then q; you can also denote it as this, p → q; and that is "p implies q."0140

Now, you could write this in a couple of different ways; you don't always have to write it "if" and "then."0151

And it is still going to be considered a conditional: back to this example, "If it is raining, then I use an umbrella."0158

If you write it without the "then," here is "then": If it is raining, I will use an umbrella; you can write it like that, too.0165

"If it is raining, then I use an umbrella" can also be "If it is raining, I will just use an umbrella."0177

You can also write it using "when" instead of the "if"; you are going to use the word when instead of if.0184

"When it is raining, then I use an umbrella": just because you don't see an if there...0192

this is still going to be the hypothesis, and then this is the conclusion.0199

You can also reword it by stating the hypothesis at the end of it: "I use an umbrella if it is raining."0205

Remember to always look for that word "if": I use an umbrella if it is raining, or I use an umbrella when it is raining.0213

Just keep that in mind: the hypothesis doesn't always have to be in the front.0222

Let's identify the hypothesis and the conclusion: the first one: I am going to make the hypothesis red, and the conclusion will be...0234

If it is Tuesday, then Phil plays tennis: well, the hypothesis, I know, is "if it is Tuesday."0247

So, "it is Tuesday" will be the hypothesis; then what is going to happen as a result?0257

Phil is going to play tennis; that is the conclusion.0264

If it is Tuesday, then Phil plays tennis.0268

The next one: Three points that lie on a line are collinear.0271

Now, this is not written as a conditional statement; so let's rewrite this in if/then form.0280

Three points that lie on a line are collinear; If three points lie on a line, then they are collinear.0290

My hypothesis, then, is "three points lie on a line"; and then, my conclusion is going to be "then they are collinear."0319

Now, notice how, when I identify the hypothesis and conclusion, I am not including the "if" and the "then"; it is following the if and following the then.0331

The next one: You are at least 21 years old if you are an adult.0339

If you look at this, I see an "if" right here; so "you are at least 21 years old," if "you are an adult."0350

Right here, "if you are an adult"--that is going to be the hypothesis; this is an example of when the hypothesis is written at the end of the statement.0359

If you are an adult, then you are at least 21 years old.0368

These examples, we are going to write in if/then form; adjacent angles have a common vertex.0379

If angles are adjacent, then they have a common vertex.0393

The next one: Glass objects are fragile; what is fragile?--glass objects.0418

So, if the objects...you can write this a couple of different ways.0424

You can say, "If the objects are made of glass"; you can say, "If these objects are glass objects..."0433

I am just going to say, "If the objects are glass, then..." what?..."they are fragile."0443

And the third one: An angle is obtuse if its measure is greater than 90 degrees.0459

If..."its"...we want to rewrite this word; if an angle measures greater than 90 degrees, then it is obtuse.0474

OK, when we are given a conditional, we can write those given statements in three other forms,0522

meaning that we can change the conditionals around in three different ways.0537

And the first way is the converse way: converse statements.0543

Oh, and then, we are going to go over each of these separately; so converse statements is the first one,0550

then inverse statements, then contrapositive statements; so just keep in mind that there are three different ways.0554

And the first one, converse statements, is when you interchange the hypothesis and the conclusion.0559

So, remember how we had if p, then q; the hypothesis is p; the conclusion is q.0566

When you switch the p and the q, that is a converse; so what happens then is: it becomes if q, then p.0571

The if and then are still the same; you are still writing the conditional; but you are just switching the hypothesis and the conclusion.0583

And when you write the converse, it doesn't necessarily have to be true.0593

It can be true or false; so again, this is going to be if q, then p.0598

And remember: our conditional statements were p to q, but then the converse is going to be q to p, q implies p, because we are switching them.0605

Here is an example: If it is raining, then I use an umbrella--that is the given conditional statement.0618

Then, the converse, by switching: this is the hypothesis; "then I use an umbrella"--that is the conclusion.0625

You are going to interchange these two; so then, "If I use an umbrella, then it is raining."0632

This is the converse statement, because you switched the hypothesis and the conclusion.0643

This is p; this is q; so then, this became q, and this is p; the converse just interchanges them.0650

Now, remember from the last section: we went over counter-examples.0665

Whenever you have some given statement, and you need to prove that it is false, then you give an example of when that statement is not true.0670

And that is when you can prove that it is false.0683

And like I said earlier, converse statements are not necessarily true; they are going to be true or false.0689

If it is true, then you can leave it at that; but if it is false, then you need to give a counter-example--an example of why it is false, or when it is false.0694

Write the converse of each given statement; decide if it is true or false; if false, write a counter-example.0704

This one: Adjacent angles have a common side.0712

Now, that is the given statement; we need to find the converse statement.0718

So, if you want to write this as a conditional (meaning an if/then statement), then you say, "If angles are adjacent, then they have a common side."0722

Then, the converse is going to be, "If"...now remember: again, you are not putting "then" first;0754

you are keeping the "if" and the "then" statement, but you are just interchanging these two;0767

so, "if angles have a common side, then they are adjacent."0773

Now, we know that this statement right here is true: "If angles are adjacent, then they have a common side"; that is true.0797

"If angles have a common side, then they are adjacent": well, if I have an angle like this; this is A...angle ABC, D...0805

this is angle 1; this is angle 2; now, I know that angles 1 and 2 are adjacent angles, and they have a common side;0824

that is the statement right here, and it is true.0842

Now, if angles have a common side, then does that make them adjacent?0847

Well, let's look at this: I see angle 2 right here, this angle, with this angle; angles 2 and ABC have a common side, which is this right here.0851

This is their common side; but they are not adjacent.0872

So, angles 2 and ABC are not adjacent angles, even though they have a common side.0877

So, that would be my counter-example; the counter-example says that this is false, because this angle right here0885

and this angle right here have a common side of BC, but they are not adjacent.0895

So, keep that in mind--that it could be false--and then give a counter-example.0902

The next one: An angle that measures 120 degrees is an obtuse angle.0911

Let's write that as a conditional: An angle that measures 120 degrees is an obtuse angle,0920

so if an angle measures 120 degrees, then it is an obtuse angle.0927

Now, we know that that is true; if an angle measures 120 degrees, maybe like that right there (this is 120 degrees), then it is an obtuse angle.0951

Let's write the converse now: If an angle is an obtuse angle, then it measures 120 degrees.0966

We know that this is true; is the converse true?0995

If an angle is an obtuse angle, does it measure 120 degrees?0999

Well, can I draw another obtuse angle that is not 120 degrees--maybe a little bit bigger?1004

This could be 130 degrees; that is still an obtuse angle.1012

So then, this right here would be my counter-example, because this is false, and I am showing an example of when the statement is not true.1016

The next one: Two angles with the same measures are congruent.1029

So, if two angles have the same measure, then they are congruent.1040

The converse (and this just means "congruent"): If two angles are congruent, then they have the same measure.1064

If two angles have the same measure...there is an angle, and here is another angle...they are the same.1102

They have the same measure, meaning that...let's say this is 40 degrees; this is 40 degrees.1113

Then, they are congruent; so if this is ABC, and this is DEF, I know that, since the measure of angle ABC1121

is 40, and the measure of angle DEF is 40, they have the same measure; then they are congruent.1138

So then, angle ABC is congruent to angle DEF; this is true.1149

If two angles are congruent, then they have the same measure; that is true, also.1163

That means that the measure of angle ABC equals the measure of angle DEF.1173

And this is the definition of congruency; so you can go from congruent angles to having the same measure; so this is also true.1183

The next one, the second statement, is the inverse statement.1199

This one uses what is called negation; now, when you negate something, you are saying that it is not that.1207

So, if you have p, the hypothesis, then you can say not p; and it is represented by this little symbol right here: this means "not p."1218

So, if a given statement is "an angle is obtuse," then the negated statement would be "an angle is not obtuse."1231

That is all you are doing; and what inverse statements do is negate both the hypothesis and the conclusion.1241

So, you are saying, "if not p, then not q"; your conditional was "if p, then q"; the inverse statement is going to be "if not p, then not q."1251

And that is how you are going to write it: like this: not p to not q; and this is the inverse.1269

Here is a given statement, "If it is raining, then I use an umbrella."1278

The inverse is going to be, "If it is not raining, then I do not use an umbrella."1281

Remember: for converse, all you do is interchange the hypothesis and conclusion.1286

With the inverse, you don't interchange anything; all that you are going to do is negate both statements, the hypothesis and the conclusion.1293

If it is not raining, then I do not use an umbrella.1299

Write the inverse of each conditional; determine if it is true or false; if false, then give a counter-example.1308

If three points lie on a line, then they are collinear.1314

The inverse is going to be, "If three points do not lie on a line, then they are not collinear."1323

We know that this right here, "If three points lie on a line, then they are collinear," is true; that is a true statement.1361

If we have three points on a line, then they are going to be collinear.1370

If three points do not lie on a line, then they are not collinear...let's see.1380

If I have a line, and let's say one point is here; one point is here; and one is right here; three points do not lie on a line.1395

Then, they are not collinear--is that true? That is true.1405

If they don't lie on a line, then they are not collinear.1409

The next one: Vertical angles are congruent.1415

If you want to rewrite this as a conditional, you can: If angles are vertical, then they are congruent.1421

The inverse statement: If angles are not vertical, then they are not congruent.1447

If angles are vertical, then they are congruent; vertical angles would be this angle and this angle right here.1474

So, they are vertical angles, and we know that they are congruent.1484

Now, if angles are not vertical, then they are not congruent.1488

Well, what if I have these angles right here?1494

They are not vertical, but they can still be congruent, if this is 90 and this is 90.1504

They have the same measure, so that means that they are congruent; so this would be false, and here is my counter-example.1511

Now, the third statement is the contrapositive; and that is formed by doing both the converse and the inverse to it.1522

You are going to exchange the hypothesis and the conclusion and negate both.1532

Remember: the converse was where you exchange the hypothesis and the conclusion; in the inverse, you negate both the hypothesis and the conclusion.1539

For a contrapositive, you are going to do both.1546

If p, then q, was just the given conditional statement; but you are going to do "if not q, then not p."1554

So, right here, we see that p and q have been interchanged; that is what you do for the converse.1563

And then, not q and not p--that is negating both: so not q to not p is the contrapositive.1573

The given statement: "If it is raining, then I use an umbrella."1585

The contrapositive: "If I do not use an umbrella, then it is not raining."1588

Here is my p; here is my q; if I do not use an umbrella...not: that means that I did negate; and this is the q statement; then it is not raining: negate p.1595

Find the contrapositive of the conditional, and determine if it is true or false: Vertical angles are congruent.1620

As a conditional, it is, "If angles are vertical, then they are congruent."1628

The contrapositive is, "If"...then I need my q, my conclusion, negated, so, "If angles are not congruent, then they are not vertical."1650

"If angles are vertical, then they are congruent" becomes "If angles are not congruent, then they are not vertical."1690

So, this is a true statement; and then, if angles are not congruent, then they can't be vertical.1700

This is also a true statement; now, for the contrapositive, when you have a conditional that is true, then the contrapositive will also be true.1709

For the converse and the inverse, it could be true or false; but the contrapositive,1724

as long as the original conditional is true, will always be true.1729

If the given conditional statement is false, then the contrapositive will be false.1736

The summary for this lesson: We have conditional statements; and this was one.1747

You write statements in if/then form, and then, from here, you can write the converse, the inverse, and the contrapositive.1762

If we know that "if p"--this is the hypothesis; the "then" statement is the conclusion; then the conditional statement is going to be, "if p, then q."1788

Or I can also write it as p → q.1807

The converse is when you switch the hypothesis and the conclusion.1815

You are going to interchange them; so it is going to be if q, then p.1822

And this is the converse: or you can write q → p; see how all they did was just switch.1833

The inverse is when you negate; you are going to use negation.1845

And this is if...back to p...back to the hypothesis...if not p, then not q.1853

And that would be like that: not p, not q; it is important to know all of these, how to write it like this and like this.1866

The contrapositive uses both: converse and inverse; and this would be if not q, then not p, not (wrong color) q, not p.1879

And another thing to keep in mind: if this is a true statement, then remember that the contrapositive is always going to be a true statement.1907

And then, for the converse and the inverse, it could be true, or it could be false.1923

This could be true or false, and this could be true or false.1928

But keep in mind that the contrapositive will be true, as long as the conditional statement is true.1932

Let's do a few examples: Identify the hypothesis and the conclusion of each conditional statement.1941

If it is sunny, then I will go to the beach: the hypothesis, again, follows the "if"; so "if it is sunny"--that is the condition--"if it is sunny."1949

The conclusion is "then"--what is going to happen as a result: "I will go to the beach."1965

The hypothesis is "it is sunny"; the conclusion is "I will go to the beach."1973

The next one: If 3x - 5 = -11, then x = -2.1980

My hypothesis is "if" this is the equation; "then" my solution is -2, and that is my conclusion.1990

Write in if/then form: A piranha eats other fish.2005

If the fish is a piranha, then it eats other fish.2014

The next one: Equiangular triangles are equilateral: so if triangles are equiangular, then they are equilateral.2040

And it is very important not to confuse the hypothesis and conclusion.2073

A keyword here is "are," or "is"; something is something else--that is a good indicator of what the hypothesis is and what the conclusion is.2080

OK, write the converse, inverse, and contrapositive, and determine if each is true or false.2092

Then, if it is false, then give a counter-example.2101

If you are 13 years old, then you are a teenager.2105

The converse is when, remember, you interchange: If you are a teenager, then you are 13 years old.2111

Is this true or false? Well, the given conditional, "If you are 13 years old, then you are a teenager," is true.2141

How about this one, "If you are a teenager, then you are 13 years old"?2151

Well, can you be 14 and still be a teenager?2156

My counter-example will be showing that, if you are a teenager, then you can also be 14 years old; you can be 15; and so on,2159

and still be considered a teenager; so this one is false.2172

The inverse is when you negate both the hypothesis and the conclusion: "If you are not 13 years old, then you are not a teenager."2178

Well, you can still be 12, and still be a teenager; so this one would be false.2210

And then, the contrapositive is when you interchange them, and you negate both.2222

If you are not a teenager, then you are not 13 years old.2232

The contrapositive is, "If you are not a teenager, then you are not 13 years old."2254

Well, if you are 13, you are considered a teenager; so if you are not a teenager, then you are not 13 years old; so this one is true.2259

And again, since this is true, then the contrapositive is going to be true.2269

Write the converse, inverse, and contrapositive, and determine if each is true or false.2280

If it is false, then we are going to give a counter-example.2286

Acute angles have measures less than 90 degrees.2290

Let me change this to a conditional statement; or, as long as you know what the conditional statement is,2300

then you can just go ahead and start writing the converse, inverse, and contrapositive.2307

Acute angles have measures less than 90 degrees; a conditional statement is "if angles are acute, then they measure less than 90 degrees."2312

And my converse is, "If angles measure 90 degrees, then they are acute."2346

Let me just write them all out: the inverse is, "If angles are not acute, then they do not measure less than 90 degrees."2379

Again, remember: inverse is when you just negate the hypothesis and the conclusion.2406

You are going to make both of them the opposites; so if angles are acute, then the inverse would be "if angles are not acute."2413

OK, and the contrapositive is when you are going to do both.2423

You are going to interchange them, and you are going to negate both.2430

"If angles do not measure less than 90 degrees, then they are not acute."2437

So, this right here, I know, is a true statement: if angles are acute, then they measure less than 90.2466

So, an acute angle is anything that is less than 90.2478

And then, if angles measure...you know, I made a mistake here: if angles measure 90...2484

sorry, it is not "measure 90," but "measure less than 90"...then they are acute.2494

Is that true--"If angles measure less than 90, then they are acute"? Yes, that is true.2504

The inverse is, "If angles are not acute, then they do not measure less than 90."2512

That is true; if it is not acute, then it is either a right angle or an obtuse angle.2519

If it is a right angle, then it measures exactly 90; and if it is an obtuse angle, then it has to measure more than 90.2525

If it is not acute, then it is not going to measure less than 90; that is true.2533

And the contrapositive is, "If angles do not measure less than 90, then they are not acute."2538

So again, if they don't measure less than 90, then they can't be acute; then it is either going to be a right angle or an obtuse angle; so this is also true.2544

And remember that, if the statement is true, then the contrapositive is also going to be true.2555

That is it for this lesson; we will see you next time.2562

Thank you for watching Educator.com!2566

Welcome back to Educator.com.0000

In this next lesson, we are going to go over some postulates that have to do with points, lines, and planes.0002

First, let's talk about postulates: what is a postulate?0011

A postulate is a statement that is assumed to be true; this is also called an axiom.0015

Postulates are accepted as fact without having to be proved.0023

Theorems are statements that have to be proved; you have to prove that it is true.0029

But postulates--we can just use them without any question if it is true or not--we don't have to prove it at all; it is just true.0035

And some postulates in your textbook--you might see that they are titled 2-2 or Postulate 2-1 or something.0046

Remember: when you name a postulate, you don't name it by that number that is used in your book,0057

because different books use different numbers, and it is in a different order.0062

If it doesn't have a name--if it just has a number, like Postulate 2.2, then remember that you have to write out the whole thing.0067

You can't just call it by the number that your book uses.0076

The first postulate that we are going to go over: Through any two points, there is exactly one line.0084

If there are any two points--I can draw two points however I want--maybe two points like that.0094

Through any two points, I can only draw one line through those two points, like that.0101

And there is no way that I can draw any other line.0110

So, if I have another two points, there is only one line that can be drawn through those two points.0113

The next one: Through any three points not on the same line, there is exactly one plane.0123

Through any three points not on the same line--meaning that they are not collinear, like that, there is exactly one plane.0130

I can only draw one plane that covers those three points--I can't draw any other plane.0140

Just like this one, through any two points, I can only draw one line--I can't draw any other type of line0148

that is going to go through those same two points--it is the same thing here.0154

Through any three points, I can only draw one single plane that is going to cover those points.0158

A line contains at least two points--"at least" meaning infinite--it contains two and a lot more.0166

So, a line contains at least two points.0178

A plane (remember, the fourth one--the next one) contains at least three points not on the same line.0189

If I have a plane, then this plane is going to contain at least three points; it is actually many, many, many--0204

but at least three points not on the same line, because if they are collinear, then it is just going to be on the same line.0213

But if they are not collinear, then it is going to be on the same plane; so this plane contains at least three points not on the same line.0223

The next one: If two points lie in a plane, then the entire line containing those two points lies in that plane.0233

So again, if two points lie in a plane (let me draw a plane, and two points lying in that plane),0244

then the entire line containing those two points lies in that plane.0256

The line that I can draw through those two points is going to also be in that plane.0262

So, if I have two points in a plane, then the line (remember: you can only draw one line through those points)0272

that you can draw is also going to be in that plane.0280

If two lines intersect, then they intersect in exactly one point.0288

If I have two lines, where they intersect is right here; where they intersect is going to be one point.0293

There is no way that they could intersect in any more than one point, because lines, we know, go straight.0304

Now, if we can bend it, then maybe it can come back around and meet again.0313

But we know that lines can't do that; it just goes straight, so their intersection is always going to be one point.0319

If two planes intersect, then their intersection is a line; so if I have (now, I am a very bad draw-er, but say I have) a plane like this,0330

and then I have a plane like this, so this is where they are intersecting; then this, where they intersect,0345

right here--that place where they are touching, where they are meeting, is a line.0361

When two lines intersect, it is going to be a point; when two planes intersect, it is going to be a line.0372

We can't just say that these two planes are going to intersect at a point,0378

because then that is not true; it is not just a single point--it is all of this right here.0384

So, it is going to be a line.0388

OK, if you want to review over the postulates again, just go ahead and rewind, or just go back and go over them again.0396

We are going to use the postulates to do a few example problems.0406

Using the postulates, determine if each statement is true or false.0411

Points A, B, and E...first of all, let's actually go over this diagram.0416

We have a plane: this is plane N; this is point A, right here; this is point B; this is C, point D; this is plane N--0423

this plane is N, right there; this point is I; this is point E.0437

So, points A, B, and E line in plane N; points A (this is point A), B, and point E (that is point E, right there) lie in plane N.0446

And we know that this is false, because E does not lie in it; A lies in it; B lies in the plane; but E does not, so this is false.0463

The next one: Points A, B, C, and E are coplanar.0480

"Coplanar" means that they are on the same plane.0488

Well, A, B...look at this...C, and E are coplanar; now, they might not be on plane N all together,0492

but they actually are coplanar, because this point...2, 3...and this one right here...I can form a plane0506

that is going to contain these four points, so this right here is true.0523

They are coplanar; it is not plane N, but they can lie on some plane, a different plane.0532

BC does not lie in plane N: here is BC right here; BC does not lie in plane N.0542

Well, it does actually lie in plane N, so this one is false, because B and C both lie (this point, B, and this point, C, lie) in plane N.0553

Remember the postulate where it says that, if two points are in the plane, then the line containing those two points also lies in the plane.0570

So, since point B and point C lie in the plane, BC has to lie in the plane.0580

Points A, B, and D are collinear: are they collinear?0589

They are coplanar, because they are all on plane N; but they are not collinear,0597

because, for it to be collinear, they have to be on the same line; and A, B, and D are not, so this one is false.0601

OK, let's go over a few more: now, we are going to determine if these are true or false.0617

Just to go over this diagram again: this right here is plane R; this right here is plane P; these are all the points.0628

Points B, D, A are part of this plane, and then, it is also part of this plane.0641

E is on plane P; H and I are not on either of them.0647

Points A, B, and D lie in plane R: is that true?0656

Here is plane R; B lies in it, D, and A; yes, it is true.0665

Points B, D, E, and F are coplanar; B, D, E, and F...well, B, D, and E are coplanar,0674

and B, D, and F are coplanar; but all four of them together--they are not coplanar.0694

So, there is no way that we can draw those four points on the same plane, so this one is false.0701

BA lies in plane P; BA, this segment right here, lies in P.0711

Well, I know that point B lies in plane P; point A lies in plane P; so the line containing those two points also has to be on that plane.0723

So, this one is true.0736

OK, we are going to go over a few more examples.0741

Use "always," "sometimes," or "never" to make each statement a true statement.0747

Intersecting lines are [always, sometimes, or never] coplanar.0752

If we have intersecting lines, no matter how we draw them (we can have them like this, or maybe like this),0760

intersecting lines are actually always going to be coplanar.0777

Can you draw a plane that contains those two points? Yes, so this one is always--always coplanar.0780

They are always going to be on the same plane.0790

Two planes [always/sometimes/never] intersect in exactly one point.0798

So, again, let me try to draw this out; I have a plane, and I have another plane...something like that.0804

Do they intersect? They intersect right here; when they intersect, are they intersecting at one point?0824

No, they intersect at a line; so this one is never: two planes never intersect in exactly one point.0834

It is always going to be a line.0848

Three points are [always, sometimes, or never] coplanar.0858

Well, if I have three points, are they going to be coplanar?0865

Yes, they are always going to be coplanar, because no matter how I draw these three points, I can always draw a plane around them.0878

Whether it is like that, or whether it is like they are collinear--they are going to be coplanar, the three points.0891

The next one: A plane containing two points of a line [always, sometimes, or never] contains the entire line.0904

A plane containing two points of a line contains the entire line--this is always.0917

As long as the two points are in that plane, the line has to also be in that plane.0934

Four points are [always, sometimes, or never] coplanar.0943

Well, this is actually going to be...let's see...if I have a plane like this, say I draw a line through that plane;0949

I can have point, point...if I have points A, B, C, and then right here, D, are all four points coplanar?0966

Now, what if I have this point right here, E? E is on this plane.0983

So, in this case, A, B, C, and E are coplanar; but A, B, C, and D are not coplanar; so this would be sometimes.0988

Two lines [always/sometimes/never] meet in more than one point.1006

Two lines, when they intersect...do they always meet at one point? Sometimes? Or never?1013

This is always at one point; can they meet in more than one point? No, so this one is never.1022

They can never meet in more than one point; they always have to meet in one point.1032

That is it for this lesson; thank you for watching Educator.com!1040

Welcome back to Educator.com.0000

This next lesson is on deductive reasoning.0002

Deductive reasoning is the process of reasoning logically--that is the keyword right here, "logically."0007

You are going to use logic from given statements to form a conclusion.0013

If given statements are true, then deductive reasoning produces a true conclusion.0019

As long as we have statements that we can show as true, then based on those statements, we can come to a true conclusion.0028

And this is the process of deductive reasoning.0044

Many professions use deductive reasoning: doctors, when diagnosing a patient's illness...0049

A few lessons ago, we learned about inductive reasoning; that is the opposite of deductive reasoning.0055

Inductive reasoning uses, remember, examples and past experiences.0061

But for deductive reasoning, each situation is unique, and you are going to look at basically facts and truths--0069

anything that is true--to come up with that conclusion.0078

Doctors, when you diagnose a patient's illness, have to look at all the facts and what is there to be able to diagnose the illness correctly.0082

You don't want doctors to diagnose based on inductive reasoning, because then, as long as you have the same symptom, then you have the same illness.0092

For example, if you have a bruise, and you go in to see the doctor, inductive reasoning would suggest that,0104

well, since the last two patients that came in with bruises had some sort of illness,0116

you will have the same illness also, just because you have a bruise.0124

But deductive reasoning...again, you have to look at each unique situation, and looking at that individual,0128

and all of the given statements, all of what is true, the facts there--using that, the doctors will diagnose the patient's illness.0137

Carpenters, when deciding what materials are needed at a worksite: each time a carpenter has a different site, they need a different material.0149

So, deciding what materials to use at that specific worksite is considered deductive reasoning.0160

Again, inductive reasoning is using examples, past experiences, and patterns to make conjectures.0174

You make conjectures; you make guesses, using "Well, it happened this way the last five times,0182

so the sixth time, I can make a conjecture that it is going to happen again."0190

So, a conjecture is an educated guess.0196

Now, with deductive reasoning, you use logic, and you use rules, to come to a conclusion.0199

With inductive reasoning, you are just kind of guessing, just by patterns, what is going to come up next.0208

But with deductive reasoning, you are actually looking at the situation, and you are going to use logic;0216

and you are going to use rules and facts to make a conclusion, to base it on something.0221

The first law of logic is the Law of Detachment.0230

Now, if a conditional is true, and the hypothesis is true, then the conclusion is true.0238

If you look at this, it will be easier to understand.0248

This is the conditional statement: if p → q is true, and p is true, then q will be true.0251

As long as the conditional is true and the hypothesis is true, then the conclusion will be true.0265

Here is an example: If a student gets an A on the final exam, then the student will pass the course.0273

That is the conditional p to q: If a student...here is p; all of this is p, "a student gets an A on a final exam."0280

Then, the student will pass the course; here is q; so p to q is true.0290

Now, David gets an A on the geometry final; here, this is this p, so that is true,0300

because the conditional statement says that if a student gets an A on the final exam, then the student will pass the course.0311

Well, David got an A on the final exam; then what can you conclude--what kind of conclusion can you make?0318

It is that David, then, will pass the course.0329

So, this conditional was true; "If a student gets an A on a final exam, then the student will pass the course"--that is the given conditional.0344

Then, David gets an A on the final exam; that is part of this.0355

So, if he gets an A on the final exam, then you can say that he is going to pass the course, because that is what the conditional says, and the conditional is true.0360

The next example: If two numbers are odd, then their sum is even.0370

Two numbers are odd--here is p; their sum is even--here is q.0380

And then, 3 and 5 are odd numbers; this is based on p--this is all based on p.0386

p → q is true, and this right here, "3 and 5 are odd numbers"...then my conclusion is that the sum of 3 and 5 is even.0394

The sum is going to be even, then, because this is the conditional.0420

If two numbers are odd, then their sum is even; and 3 and 5 are odd numbers; then, the sum of 3 and 5 is even.0424

You are using the conditional and a hypothesis; then you are going to come to a conclusion.0433

And this is the Law of Detachment: if p → q is true, and p is true, then q is true.0442

The next one is the Law of Syllogism; this one is very similar to the transitive property of equality.0454

If you remember, from Algebra I, you learned the transitive property.0461

The transitive property says that, if A equals B, and B equals C, then A equals C.0466

If A equals B, and B equals C, then, since these two are equal, A equals C.0475

This is very similar to that: the Law of Syllogism says that if the conditional p → q is true,0491

and q → r, that conditional, is true, then p → r is also true.0502

So then, here you have two different conditional statements.0511

You have p → q, and then you have q → r; now remember, this q and this q have to be the same.0515

p → q is true; q → r is true; this is a different conclusion; then, this hypothesis, p, to this conclusion, r, is going to also be true, just like this one.0522

A to B and B to C...then A is equal to C.0537

Let's just do a couple of examples: Using the two given statements, make a conclusion, if possible.0544

If M is the midpoint of segment AB, then AM is equal to MB.0554

If I have segment AB, and M is the midpoint (this is M), then AM is equal to MB.0563

If the measures of two segments are equal, then they are congruent.0580

Here, this segment and this segment are equal; right here, that is what it says.0602

Here is AM, and here is MB, and they are equal to each other; then, they are congruent.0609

So, all of this right here--this is all p; this first one would be p → q; and then, this is q.0616

This next one, "The measures of two segments are equal," is saying the same thing as this right here: AM = MB/two segments are equal.0636

So then, this is using q; then they are congruent--now, this is a new conclusion, so this is r; so this is q → r.0651

So, my conclusion...see, right here, the Law of Syllogism says p → q; there is p → q;0668

then q → r--this q → r; then p → r is also true.0674

So, I can come up with a true conditional statement by using this.0679

Then, I can say that my p is here; so if M is the midpoint of segment AB, then the segments are congruent.0686

And I can also say "then AM is congruent to MB," because this one uses AB, so I can just say AM is congruent to MB.0728

I can write it like this, or I can write it like this.0744

Here, I used p; all of this is p; and then, the segments are congruent, so that is r; so this was p to r.0747

So, since this is true, and this is true, then this is what I can conclude: p → r is also true, by the Law of Syllogism.0764

Let's do the next one: If two angles are vertical, then they do not form a linear pair.0776

Here is p; then they do not form a linear pair--this is q; this one is p → q.0787

Then, if two angles are vertical--look at this one--this is the same as right here; so this one is p.0801

Then, they are congruent; this is r; so this is p → r.0809

Well, here I have p → q; and the Law of Syllogism says that p → q and q → r have to be true.0819

I can't have p → q and then p → r; I can't come up with a true conclusion, because here it is not q → r; it is p → r.0828

With this, I can't form a conclusion; so this one is no conclusion.0840

Let's do a few examples: we are going to use the laws of logic, the ones that we just learned, the Law of Detachment and the Law of Syllogism.0853

And we are going to determine if statement 3, the third statement, follows logically from true statements 1 and 2.0865

Based on the first one and the second one, we are going to see if the third one is going to be a true conclusion.0874

Number 1: Right angles are congruent--that is the first statement.0883

Now, this is not written as a conditional; so if you want, you can rewrite it as a conditional.0889

Or you can just remember that this part right here is going to be the hypothesis, and this part right here is going to be the conclusion.0894

I will just write out the conditional of "right angles are congruent": "If the angles are right angles, then they are congruent."0904

And that is the congruent sign; if angles are right angles, then they are congruent.0926

Now, it is easier to see that this is my hypothesis; that is p; and "they are congruent"--this is q.0931

This first one was p → q; now, the second statement is "Angle A and angle B are right angles."0942

Here we have right angles; now, do we see that?--that sounds familiar to me.0954

It is right here; the angles are right angles; so angles A and B are right angles.0961

This is p; or we can write it here--p.0967

Then, angle A is congruent to angle B: is that the correct conclusion?0975

Well, here, if the angles are right angles, and it says that angle A and angle B are right angles, then what?0982

They are congruent; so then, this says that they are congruent; so this is q.0990

This is true; this is a valid conclusion, based on the Law of Detachment, because the Law of Detachment says that,0996

if p → q is true, and p is true, then q is true.1009

So, it is valid; this is the Law of Detachment.1013

The next one: Vertical angles are congruent.1023

Vertical angles: this one is p; them being congruent: that is q, so this one is p → q.1031

Angle 1 is congruent to angle 2--now, is that from p or q?1043

That is from q, because it says that angles are congruent here; so this one is q.1051

And angle 1 and angle 2 are vertical angles--this is p...this is actually...I wrote p instead of q right here.1058

This one is q, and the conclusion was that angle 1 and angle 2 are vertical angles, which is p.1069

Now, we don't have a law of logic that says that if p → q is true, and q is true, then p is true.1078

That is not any law; it looks like the Law of Detachment, but the Law of Detachment is that if p → q is true, and p is true, then q will be true.1088

OK, so in this case, this is an invalid conclusion.1103

See, p → q and p--then q will be true; it can't be the other way around.1114

This is invalid; this is actually the converse, and that is not true.1120

Again, using the Law of Detachment and the Law of Syllogism, determine if statement 3 follows logically from true statements 1 and 2.1131

And state which law is used.1140

The first one: inline skaters live dangerously: here, "inline skaters" would be p; they "live dangerously"--that is q.1144

"If you live dangerously"...that is the same thing as q; so this is q..."then you like to dance"; this is a new statement, so this is r.1162

"Inline skaters"--this is p--"like to dance"--this is r.1178

So, here "inline skaters live dangerously" is p → q; this is q → r; then the third statement,1188

"inline skaters like to dance"--this is p → r; and this is valid by the Law of Syllogism.1199

This one says, "Inline skaters live dangerously"; that is p → q.1216

If you live dangerously--that is the same statement as this one right here--then you like to dance; that is q → r.1223

So, this one right here and this one right here are the same.1232

"Inline skaters"--that same statement right there is p--"like to dance"--that is r.1237

And the Law of Syllogism, remember, says if p → q is true, and q → r is true, then p → r is true.1244

So, it is like the transitive property--the Law of Syllogism.1251

The next one: "If you drive safely, the life you save may be your own."1256

Here, this is p; "the life you save may be your own"--here is q.1266

"Shani drives safely"--that is from p; "the life she saves may be her own"--this one is q.1276

This is the same as this one; so the first statement is p → q, and the next one is p; the conclusion,1291

"the life she saves may be her own," is q; so based on 1 and 2, based on these two, we are able to get this.1304

Yes, this is valid; and this is one is by the Law of Detachment.1314

The Law of Detachment says that if p → q is true, and p is true, then q is true.1325

We are going to do a few more examples: the first one: Draw a conclusion, if possible; state which law is used.1335

If you eat to live, then you live to eat: "If you eat to live"--this one is p--"then you live to eat"--that is q; that is p → q.1344

"Christina eats to live": that is from statement p, so draw a conclusion.1362

Our conclusion is, then, "Christina lives to eat," because if p → q is true, and p is true, then I can conclude that q is true.1372

And this one was by the Law of Detachment.1393

The next one: "If a plane exists, then it contains at least three points not on the same line."1404

"If a plane exists"--there is p--"then it contains at least three points not on the same line"--there is q.1412

And to draw this out: this is just saying that if I have a plane, then contains at least three points in the plane that are not on the same line.1423

Plane N (let me draw plane N--here is plane N) contains points A, B, and C, which are not on the same line.1440

If a plane exists, then it contains at least three points not on the same line.1466

Plane N contains points A, B, and C, which are not on the same line.1471

Well, all of this right here is from statement Q; we have...so I have p → q, and then I have a q.1477

So, I cannot come to a conclusion; I cannot draw a conclusion, because there is no law that says that if p → q is true, and q is true, then p is true.1507

So, I can't say, "Then plane N exists"--that is not a conclusion that I can come to.1521

In this case, my answer will be no conclusion--it cannot be done.1527

Draw a conclusion, if possible; state which law is used.1541

If you spend money on it, then it is a business; if you spend money on it, then it is fun.1546

Let's label these: this right here is p; "then it is a business" is q; so this is p → q.1552

"If you spend money on it"--well, that is p; "then it is fun"--this is r; it is not the same as q, so it is r; this is p → r.1563

Now, can I draw a conclusion based on these statements?--no, because there is no law that says that,1579

if p → q is true, and p → r is true, then q → is true.1591

It has to be p → q; so the Law of Syllogism says p → q and q → r; then p → is true--this is the Law of Syllogism.1600

So, in this case, since it is p → q and p → r, this has no conclusion.1627

The next one: if a number is a whole number, then it is an integer.1640

Remember that whole numbers are numbers like 0, 1, 2, 3, and so on.1645

And integers are whole numbers and their negatives, so it is going to be -2, -1, 0, 1, 2, and so on; those are integers.1656

"If a number is a whole number"--there is my p--"then it is an integer"--there is a q.1672

"If a number is an integer"--isn't this q?--"then it is a rational number"--this is r.1684

And rational numbers are numbers that are integers (it could be -2); I can have fractions; I can have terminating decimals--all of that.1696

Now, remember: these given statements are true statements, and you are trying to see if you can use those true statements to draw a conclusion.1714

Here is p → q; this one is q → r; remember: if we have p → q and q →, then we can say that p → r is true.1727

That is going to be our conclusion: p → r; so I can say, "If a number is a whole number, then"--here is p--1737

I am going to draw my conclusion, p → r--"then it is a rational number."1759

All of this is p, and all of this is r; and that would be valid because of the Law of Syllogism.1771

The next example: Determine if statement 3 follows logically from statements 1 and 2; if it does, state which law is used.1791

Based on 1 and 2, we are going to see if number 3 is valid.1800

If you plan to attend the university of Notre Dame, then you need to be in the top 10% of your class.1806

Here is my p; "you need to be in the top 10% of your class"--there is my q; so this one is p → q.1823

Jonathan plans to attend Notre Dame; so this one is p.1836

Jonathan needs to be in the top 10% of his class; this is q--yes, that is q--"then he needs to be in the top 10% of the class."1848

Based on these two, numbers 1 and 2, these are true statements; statement 1 is true, and statement 2 is true.1861

Then, is my conclusion, my statement 3, true?--yes, this statement is valid, because of the Law of Detachment.1868

The next example: Determine if statement 3 follows logically from statements 1 and 2.1889

If it does, state which law was used.1896

We are going to see, again, if the third statement is valid or invalid, based on these two true statements.1899

So, if an angle has a measure less than 90, then it is acute.1908

"An angle has a measure less than 90"--that is my p; then "it is acute"--this is q; so my conditional is p → q.1915

"If an angle is acute"--well, isn't that what this is right here?--so here is q.1927

"Then its supplement is obtuse"--the supplement is an angle measure that makes it 180.1936

So, if we have two supplementary angles, then it is two angles that add up to 180.1950

A supplement of an angle would be the number, the angle measure, that you would have to add so that it would add up to 180.1955

Then, it is obtuse; this is r; this is the new statement, so here we have q → r.1963

"If an angle has a measure less than 90"--here is my p; all of this is p--"then its supplement is obtuse"--this is r.1972

So, this, my third statement, was p → r; well, does that follow any rule, any law?1996

p → q is true; q → r is true; then p → r is true; so this is valid, and it is from the Law of Syllogism.2006

And the next example: If a figure is a rectangle, then its opposite sides are congruent.2027

If I have a rectangle, its opposite sides are congruent; so this is congruent to here, and this is congruent to here.2035

AB is congruent to DC; so if I have ABCD, AB is congruent to DC, and AD is congruent to BC.2051

The figure is a rectangle--there is my p; then its opposite sides are congruent, so there is q.2067

AB is congruent to DC, and AD is congruent to BC.2075

Here is q, because it says "if the opposite sides are congruent"; ABCD is a rectangle; this is p.2092

I am going to use a different color for that one; this is p.2111

So, statements 1 and 2 are p → q, and statement 2 is q, and my third statement is p.2116

Can you use these two to make this conclusion, that p is true?--no, so this is invalid.2129

This statement right here is invalid; the Law of Detachment says, if p → q is true, and p has to be true, then q is true, not the other way around.2136

This one is invalid; make sure that this second statement has to be p, and then your conclusion is going to be q.2148

OK, well, that is it for this lesson; thank you for watching Educator.com.2160

Welcome back to Educator.com.0000

For this lesson, we are going to talk about some properties of equality, and we are going to work on some proofs.0002

Going over some properties first: these are all properties of equality,0013

meaning that they have something to do with them equaling each other, something to do with the word "equal."0017

Now, the first one, the addition property of equality, is when you have, let's say, numbers a, b, and c.0028

If a equaled b, if the number a is the same as b, then if you add c to a, then that is the same thing as adding c to b.0039

So, if a = b, then a + c = b + c; that is the addition property of equality, because you are adding the same number to a and b, since a and b are the same.0051

And the subtraction property of equality: again, you have numbers a, b, and c.0070

a is equal to b; then, a - c is equal to b - c, as long as you subtract the same number.0076

But when you are dealing with subtraction, then it is the subtraction property.0087

But as long as you are subtracting the same number from both sides, then it is still the same.0090

You still have an equation, with equal sides.0096

The multiplication property of equality: again, you have numbers a, b, and c.0101

If a is equal to b, then a times c is equal to b times c; so again, you are multiplying the same number.0106

And the division property: for the numbers a, b, and c, if a is equal to b, then a/c is equal to b/c.0115

Now, here you have to look at this c, because you are dealing with division; so this can also be a/c = b/c.0132

This, even though it is a fraction, also means a divided by c; and when you are dealing with that,0143

since c is now the denominator, we have to keep in mind that c cannot be 0, because we can't have a 0 in the denominator.0150

So, be careful with that.0161

Now, I want to go back over these again; and since the next couple of lessons, we are going to be talking about segments0163

and angles, if I have, let's say, the measure of angle 1, the measure of angle 1 equals the measure of angle 2.0172

So then, if the measure of angle 1 is representing a, and the measure of angle 2 is representing b,0185

then the measure of angle 1, plus the measure of angle 3 (c is a new one) equals...what is b?...0191

the measure of angle 2, plus the measure of angle 3.0200

So, this is also the addition property of equality, but just using angles now.0204

The measure of angle 1, plus the measure of angle 3, equals the measure of angle 2, plus the measure of angle 3.0209

Then, you are adding the same angle measure to both of these sides.0214

The subtraction property is the same thing: if I have, let's say, the measure of angle 1,0220

minus the measure of angle 3, then that is the same thing as the measure of angle 2, minus the measure of angle 3.0230

The multiplication property does the same thing, and the division property would also be the same thing.0240

The reflexive property of equality: this one is when you have one number, a;0252

for every number a, then a equals a--it equals itself; a = a is the reflexive property.0261

You can have a segment AB equaling itself, AB; this is also the reflexive property; measure of angle 1 = measure of angle 1--reflexive property.0271

When you write this, you can write this like "reflexive"...we can write "property"...0286

and for the equality properties, even the ones that we just went over, the addition property,0296

subtraction, multiplication, and division--since they are all properties of equality,0301

you can write "reflexive property of," and then you can write an equals sign next to it, like that: "reflexive property," and then an equals sign.0306

And that equals sign represents the type of property that it is.0314

So, it is the reflexive property of equality.0321

The symmetric property is different than the reflexive property, because you are given two numbers,0326

a and b; you are saying that a equals something else; if a = b, then...and then, you are just going to flip it; and you say b will then equal a.0333

So, if AB = 10, then you can say 10 = AB; and that is the symmetric property.0348

For the symmetric property, you can just write "symmetric property of equality" like that, too.0361

The transitive property of equality: For all numbers a, b, and c, if a = b, and b = c, then a = c.0370

So, let's use angles: if the measure of angle 1 equals the measure of angle 2,0382

and the measure of angle 2 equals the measure of angle 3, then since this and this are the same,0390

the measure of angle 1 equals the measure of angle 3.0401

If this equals that, and that equals something else, then these two will equal each other; and that is the transitive property of equality.0404

This one you can write as "trans. property of equality" for short.0414

A couple more: the substitution property of equality: whenever you replace something in for something that is of the same value,0427

then you are using the substitution property; so if you have numbers a and b, and if a = b,0441

then a may be replaced by b in any equation or expression.0447

If I tell you that x = 4, and x + 5 = 9, then I can take this; since x is equal to 4, I see an x here;0452

so since this and this are the same, I can just replace the 4 in for x...plus 5, equals 9.0475

I am substituting in this for this; and that is the substitution property.0485

For the substitution property, be careful not to just write "sub.," because this can also be the subtraction property.0495

You can just maybe write it like that, or maybe you can write the whole thing out: "substitution property of equality."0504

The distributive property of equality: for all numbers a, b, and c, a times the sum of b and c is equal to ab + ac.0515

Remember: you take this value right here; you multiply it to all the values inside.0527

So, it is going to be a times b, and then plus a times c; and that is the distributive property.0536

You can also go the other way; you can take it from here; you can factor out the a.0548

We have an a in both terms; factor it out; in this term, I have a b, plus...and in this term, I have a c left; that is also considered the distributive property.0553

For the distributive property, you can write it like that; you can write "distributive of equality"; you can write "prop."0565

These are all properties of equality; we are going to be using them pretty often in what is called a proof.0578

And there are a couple of different types of proofs, but the main one is called the two-column proof.0587

And a two-column proof is just a way of organizing your reasoning, and it is deductive reasoning.0594

When you use two-column proofs, you use them to show how to come up with some kind of conclusion.0605

Remember: with deductive reasoning, we talked about having some true statements,0617

and using facts and different definitions and so on to come up with a conclusion.0623

And a two-column proof is just a way of organizing those things.0632

For a two-column proof, you are going to have a given statement, and the given statement is just whatever is given to you,0638

the information that is given; and it can be maybe the values of angles, or the values of the side measures--whatever.0652

Whatever they give you, whatever is given to you, is going to go right here, as given.0662

That is going to be given, and then they are going to give you a "prove" statement, what to prove.0668

Given this information, your conclusion--how will you get to this right here?0677

They are going to give you both statements; and then, on this side, you are going to have a diagram or some kind of drawing,0682

some kind of picture of this proof--some kind of drawing, maybe a diagram; that is going to go right here.0693

And then, right below it, you are going to have something that looks like this.0709

And it is a two-column proof, so you are going to have two columns.0719

On this column, you are going to have statements; on this column, you are going to have reasons.0722

You are going to state something--you are going to state your facts, your different things.0733

And then, on this side, you are going to have reasons for that statement.0739

You can't just say something--you have to have a reason; you have to back it up with something.0744

Why is that statement true? You are going to do numbers 1, 2, 3, 4...and it is going to go on.0748

And then, your reasons: 1, 2, 3, 4...you have to have a reason for every statement you write down.0756

Now, any time you do a two-column proof, the first statement is always going to be your given statement.0763

Whatever is written here, you are also going to write here.0772

You are going to start with your given; and then, your last statement...however many...0776

Now, you don't always have to have 4 or 5; it is usually going to be around 4, 5, or 6, but you can have less; you can have more.0784

It depends on the proof; but your last statement is going to be this statement.0793

Whatever is written here is going to be your last statement.0801

And then, for number 5, you are going to have a reason for that statement.0804

This is what a two-column proof looks like; now, if you are so confused by what a two-column proof is, think of directions.0808

From your house to, let's say, school, or from your house to a friend's house, you have a starting point.0821

You are starting at some place, and you are going to head over to school, or your friend's house, wherever it is.0833

You have directions; if you are to give someone directions to school from your house--or anywhere--0843

the starting point...you have point A to point B; you have steps to get from point A to point B.0850

If you are at home, how are you going to get to point B?0860

You make a right here, make a left here, or whatever it may be; you have directions.0864

There are steps to get there; this is exactly the same thing.0869

The given statement...this is where you start; that is point A--that is your starting point; that is like your house.0874

This statement right here, the "prove" statement, is point B--that is where you have to end up at; that is your destination.0882

You have to go from point A to point B; but again, you can't just snap your fingers and get there.0892

You have steps; you have directions to get there.0901

For each (maybe "make a right turn"; "make a left turn here"), you can't skip any steps, because it has to lead from point A,0905

and then through all of these steps, you are going to end up at point B.0916

And that is what a two-column proof is; they are just saying how you get from here to here.0924

And all you have to do is list out your statements: the starting point, point A, is going to be on line 1, statement 1.0928

Your last statement is going to be right here, your "prove" statement.0937

And you are just going to have reasons for that: why is this statement true? Why is this statement true?0941

Now, when you write your given statement for step 1, your reason is always going to be "Given."0946

That is the reason; this statement is true because it is given--that is given to you.0956

So, step 1 is this part right here, the given statement; and the reason for that is "Given."0964

So, here is an example of a proof: now, here, the statements are just listed, and the reasons are just listed.0975

It is a two-column proof; you can draw a line out like this and draw a line down like this.0983

Or you can just do it like this; as long as you have two columns, a column for statements and a column for reasons, you still have a two-column proof.0992

Again, here is your given statement; you have a few things that they give you; and prove this.1002

So, look at the statements: now, for step 1 (they are not all listed out, so let me write them out here),1011

AC is 21; that is this right here; now, you have to write out all of them.1026

So then, see how only this is written out; so I am going to write in the other ones.1031

AB = 2y, and BC = 3y - 9; those are all of your statements.1037

Now, here, AB is 2y; so I can write that in; so use this diagram to help you get from here, point A, to point B.1050

Write it in: AB is 2y; BC is 3y - 9; and then, AC, the whole thing, is 21.1065

And, given this information, they want you to prove that y equals 6.1080

Step 1: All that I did was to copy down all of the given statements right there.1088

And the reason for that is "Given."1095

Now, the next step: AB + BC = AC...well, that is because, since I have AB, and I have BC, and I have AC,1102

AC, the whole thing, is 21; but since I need to solve for y, I need to look at where my destination is.1119

Where am I trying to get to?--to what y is--my value for y.1127

Well, y, I see here, is from AB, and from BC, not from AC.1131

So, how do I mention these parts, these segments, in relation to the whole thing?1139

This is part of the segment; AB is part, and BC is another part, of this whole segment, AC.1149

If you remember, from chapter 1, we talked about segments, and then their parts.1160

I can say that AB + BC = AC; this part, plus this part, equals the whole thing: AB + BC = AC.1170

And the reason for this step, this statement, AB + BC = AC: if you remember, that is called the Segment Addition Postulate.1186

And you can just write it like this for short: the Segment Addition Postulate.1206

Now, why did I write this down--why is this step here?1214

It is because I know that, in order for me to find the value of y, I have to look at these parts, AB and BC.1218

I can't just look at the whole thing; so when I have to look at the parts, compared to the whole thing,1227

then I have to use the Segment Addition Postulate.1233

And then, what happened here? The next step: 2y + 3y - 9 = 21; so how did I get from this step to this step?1237

What happened here? Well, I know that AB is what?--AB is 2y; BC is this; so, guess what happened right here.1252

You see that...and 21; AC is 21; so, since AB is 2y, just replace AB for 2y, and then replace this for this, and replace this for this.1272

Whenever you do replacing, whenever you replace something for something else, in an equation or expression, that is the substitution property of equality.1292

Now, remember: be careful not to write "sub." because that can mean subtraction; "substitution" is the shortest you can write it.1313

Or you can write the whole thing out.1323

The next step: from here, 5y - 9 = 21--well, how did you get from this step to this step?1327

You did this plus this; you just simplified it, and more specifically, you added; so this would be the addition property (and this is for the "equality").1340

Now, here, number 5: you did 5y, and then you added 9 to both sides; this is the addition property, because you added.1364

And then, from here, how did you get from this step to this step?1385

You divided by 5 on both sides; so this is the division property of equality.1391

And then, since we have this statement, which is the same as this statement right here, we have arrived to our destination, to point B.1406

And once you do that, then you are done; so as long as you start here and you end up here, then you are done.1416

The next example: the measure of angle CDE (there is angle CDE) and the measure of angle EDF are supplementary.1431

Prove that x = 40.1442

So, here I am going to write in...now, for this one, this is x, and this is 3x + 20.1446

Sometimes, they give you the information on the diagram; they might not always give it to you in the given.1466

The given is very important, but you have to look at the diagram, too, because they might label something--1472

an angle, or give you some measure or length, and they might just write it in the diagram.1478

So, that is very, very important to have; if there is no diagram, then draw one in, because that is going to help you.1485

Especially if you are very visual--if you are a visual learner--then you should draw it in and write in whatever is given to you.1491

It will help you with your steps.1502

Number 1: Angle CDE and angle EDF are supplementary.1506

Now, to review over supplementary: supplementary means that two angles add up to 180 degrees.1514

These two angles right here, angle CDE and angle EDF, form a linear pair, meaning that,1529

when you put them together, they form a line; see how there is a line right there--so they are a linear pair.1545

And linear pairs are always supplementary, because a line measures 180 degrees.1555

So, if you have two angles that form a line, then they are supplementary.1563

If you look at supplementary angles, supplementary angles are just any two angles that add up to 180.1571

So, supplementary angles don't always form a linear pair; sometimes they do; sometimes they don't.1576

If you just have two angles that are separated, then they don't form a linear pair, but they can still be supplementary.1582

On to our proof: the reason for #1 is "Given."1591

And then, #2: I want to find x, so if I know that these two angles are supplementary, meaning that they add up to 180,1600

and then I know that these two angles together add up to 180, then I can find x that way.1614

But then, there are steps that I need to take to get there.1621

The next step is going to be that the measure of angle CDE, plus the measure of angle EDF, equals 180.1625

Now, we know that, since it says "supplementary," I can just say, "Well, since they are supplementary, then I add them together, and they equal 180 degrees."1640

And that is because of the definition of supplementary angles.1651

The definition of supplementary angles says that, if two angles are supplementary, then they add up to 180.1662

Any time you go from something supplementary to then making them add up to 180, then that would just be the definition of supplementary angles.1674

Any time you do this step, the reason will be "definition of supplementary angles."1683

The next step: now that I gave these two angles, adding them up to 180, now I have to use x somewhere, because I need to prove that x equals 40.1692

So, this angle right here became x, and then the measure of angle EDF is 3x + 20.1709

So, what happened here? Instead of writing this one, you wrote x; and instead of writing this one, you wrote 3x + 20.1722

Step #3: Since you replaced something, that is the substitution property of equality.1731

Now, #4: This right here, in the last proof--this could be the addition property, because you are adding it.1746

But it could also be the substitution property, because you are just substituting in these two for this value.1759

Let's just write "substitution property of equality."1766

And then, #5: To get from here to here, you subtracted 20 from both sides, so #5 is going to be the subtraction property.1773

And just so that you don't get confused, you can write it out, or you can just write "subtraction property," or "subtract. of equality."1798

In the next step, you divide it by 4 to get x = 40, and that is the division property, because you divided.1808

And then, we know that this is the final step, because that is what that is.1821

OK, another example: The measure of angle AXC and the measure of angle DYF...1831

oh, this is supposed to be written as "equal"; so then they are equal.1848

The measure of angle AXC and this angle are the same; and the measure of angle 1, this one, is equal to the measure of angle 3.1858

And by doing this, this is showing that they are the same.1872

So, if I do this one time, and I do this one time, that means that they are the same.1876

And I have to prove that the measure of angle 2 is equal to the measure of angle 4.1883

For this one, I don't have any statements, so we are going to have to do the statements on our own, and then come up with the reasons as we go along.1892

#1: I am going to write that the measure of angle AXC equals the measure of angle DYF,1899

and that the measure of angle 1 equals the measure of angle 3.1913

OK, and my reason for that is "Given."1925

Now, my next step: since I know that I am trying to prove this and this, that these two are equal,1932

I need to break down this big angle into its parts.1948

I know that angle AXC equals the measure of angle 1 plus the measure of angle 2.1959

So, let me write that out: the measure of angle 1, plus the measure of angle 2, equals the measure of angle AXC.1968

And the reason why I do that is because I need to somehow get that angle 2 in there somewhere.1983

And I know that these equal each other, and I know that the measure of angle 1 and the measure of angle 3 equal each other.1992

How am I going to come up with angle 2?1999

I can say that this one, plus this one, equals this big thing.2006

I am getting it in there somehow: the measure of angle 1, plus the measure of angle 2, equals the measure of angle AXC.2013

Now, I am going to do the same thing for this one, in the same step: the measure of angle 3 plus the measure of angle 4 equals the measure of angle DYF.2018

And the reason for that, if you remember from Chapter 1: this is the Angle Addition Postulate.2040

And my third step: Since all of this equals this, and all of this equals that, look at my first step right here.2060

I know that they equal each other; well, if these equal each other, doesn't that mean that all of its parts equal each other?2077

If this big angle and this big angle equal each other, then angles 1 and 2 together equal angles 3 and 4 together.2085

So, my next step is going to be: The measure of angle 1, plus the measure of angle 2,2095

equals the measure of angle 3, plus the measure of angle 4, because all of this right here equals AXC,2103

and all of this right here equals the measure of angle DYF.2114

And they equal each other; that means that all of this and all of this equal each other; so that is the only step right there.2118

Step 3: I basically just used this right here, and I substituted in the parts for that.2126

So, step 3 is going to be the substitution property; and I can put "equality."2135

Then, for #4: Now, always keep in mind what you have to prove.2146

I have to prove that this one equals this one; so I have to somehow get rid of this and this.2156

Now, look back at step 1; if you look back at step 1, see how the measure of angle 1 equals the measure of angle 3.2167

Well, here is the measure of angle 1, and here is the measure of angle 3.2177

So, since they are the same, I can use the substitution property to replace...2180

maybe measure of angle 3 for this, or measure of angle 1 for that, because they are the same.2186

I am going to just substitute in the measure of angle 1 in place of 3, since they are the same.2194

Do you see that? This is the measure of angle 1, in place of the measure of angle 3, since they are the same, since they equal each other.2209

And my reason for this one is, again, the substitution property; this property is actually used quite often.2218

And then, here, since these are the same, I can just subtract it out.2230

Then, these cancel out; I get that the measure of angle 2 equals the measure of angle 4.2238

My reason is the subtraction property of equality.2247

And I know that I am done, because this stuff is the same as that stuff.2257

Now, I know that this seems really long; but once you get used to it, and once you get more familiar with proofs,2262

they actually become kind of fun, and it is not so long; it is not so bad.2272

It is just that, since we are going over each step, and we are going over each reason, it just seems a lot longer than it is.2276

These next few examples...we are going to just go over the properties that we went over.2289

Name the property of equality that justifies each statement.2304

If 5 = 3x - 4, then 3x - 4 = 5: well, this one right here...remember when we had the property "if a = b, then b = a"?2309

This property is the same property as if I said that if ab = 10, then 10 = ab.2334

And this is the symmetric property of equality, meaning that it is the same on both sides; so you flip it, and it is the same.2347

The next one: If 3 times the difference of x and 3/5 equals 1, then 3x - 5 = 1.2360

So, what happened here--how did you get from this to this?2371

Well, it looks like this was distributed over to everything; this became 3 times x, which is 3x; and 3 times this; 3 times 5/3...2375

this is over 1; you can cross-cancel that out, and that becomes 5.2391

Then, that is how you got that 5; and minus 1; this was the distributive property of equality.2397

You can just also write "distributive...equality."2411

Name the property: Here, this is written: if 2 times the measure of angle 180...2417

no, if 2 times the measure of angle ABC equals 180 (that is how it is supposed to be written), then the measure of angle ABC equals 90.2439

OK, so if 2 times this angle is 180, then if you solve out for the measure of angle ABC, you get 90 degrees.2456

And so, how did you get from this step to that step over there?2467

Well, it looks like you divided the 2; and the measure of angle ABC equals 90; so this one was the division property of equality,2472

because you divided the 2 to get the answer--divided 2 into both sides.2487

Name the property of equality that justifies each statement.2497

For xy, xy = xy; well, this one right here--if something equals itself, this is different than the symmetric property.2501

The symmetric property is when you have something equaling something else.2511

And then, you can reverse it and say that the second thing equals the first thing.2516

In this one, there is no second thing; it is just one thing, and that one thing equals itself, so a = a; apple = apple; xy = xy.2523

Any time you have that, it is the reflexive property; "reflexive," or "reflexive property," and this is used quite often, too, in proofs.2537

If EF = GH, and GH = JK, then EF = JK.2554

Well, if 1 = 2 and 2 = 3, then 1 = 3; this is the transitive property of equality.2563

The next one: if AB + IJ = MX + IJ, then AB = MX.2585

What happened from here to get that? It looks like this happened: this is the subtraction property of equality.2595

The next one: if PQ = 5, and PQ + QR = 7, then 5 + QR = 7.2611

So, this was the equation; there is a value of PQ, and then 5 was replaced for PQ; this is the substitution property of equality.2621

Be careful when you are writing "subtraction" and "substitution."2640

It would probably be best to just write out the whole word.2644

But if you are going to write it like this, then make sure it is obvious what you are writing--subtraction property or substitution property.2648

And that is it for this lesson; we will work on some proofs for the next lesson.2660

So, we will see you next time--thank you for watching Educator.com.2666

Welcome back to Educator.com.0000

This next lesson is on proving segment relationships.0002

From the previous lesson, the concept of proofs was introduced.0007

And for this lesson, since we are going to be proving segment relationships, we are going to try and start setting up proofs and get you more familiar with them.0014

First, let's talk about what makes a good proof: a good proof is made of five essential parts.0031

First, state the theorem to be proved--you are going to state what you are going to prove.0041

List the given information--you are going to list everything that is given; make sure that it is good information and that it is all listed.0055

#3: If possible, draw a diagram to illustrate the given information--again, a diagram or some kind of picture or drawing0064

to show what you are dealing with, the statements and stuff.0075

State what is to be proven: you have a given statement, and you have a "prove" statement.0082

Use a system of deductive reasoning to complete the proof--you are going to go from point A to point B to complete the proof.0087

We have, on the two-column proof, the first column, which are all statements, and the second column, which are all reasons.0101

So, some of the reasons that you can list under that column can include undefined terms--just terms that you have previously learned;0107

definitions--this includes any type of definition of the terms, like the definition of...and you would just write definition as def. for short...0127

of maybe congruent segments, or maybe the definition of supplementary angles, and so on.0141

When you use definitions, you are just going to write "definition of," and then whatever it is.0157

Postulates: an example of a postulate would be the Angle Addition Postulate; so you can write Angle Addition Postulate.0163

And for most of these, you can make them shorter; if there is no name for it, then you can just write out the statement in a short way--0177

like, instead of writing out "congruent," you can just write this sign; instead of writing the whole word "definition," you can just write "def."0189

For the Angle Addition Postulate, it would just be Angle Addition Postulate; that is one postulate that you can use in your reasons.0196

Previously proven theorems: Theorems, unlike postulates...0205

Remember: with postulates, we talked about how they don't have to be proven; we can just assume them to be true.0212

So, once a postulate is introduced, we can just go ahead and start using them for our reasons, if we need to.0218

Theorems, however, have to be proven; once the theorems are proven (and they are usually proven in your textbooks), you can use them.0224

Once they are proved to be true, then theorems can also be used in your reasons.0238

Here is a proof on the congruence of segments.0252

Remember: we talked about, last lesson, our properties of equality.0259

We talked about the reflexive property, the symmetric property, and the transitive property.0265

We talked about these properties, but they are of equality; so they are good only when you are using the equals sign for them.0270

The reflexive property, remember, was that if you have a number a, then a = a; it equals itself--that is the reflexive property of equality.0278

The symmetric property is when you have that if a = b, then b = a; you flip them, and then that is the symmetric property of equality.0289

The transitive property was when you do that if a = b and b = c, then a = c; and that is the transitive property of equality.0299

So, these three properties can also be used for congruent segments.0309

But congruence is a little bit different; when you talk about segments that are congruent, it is a little bit different than equal.0317

If you are going to use them for segments, to show their congruence, then you can't write "reflexive property of equality."0328

So, instead of writing "reflexive of equality," like you did for the last lesson,0338

if it is for equality, you are going to write it as "reflexive property of congruent segments."0345

So, depending on how you use the reflexive property, you are going to write it in different ways.0359

If you are going to write it using an equals sign, then you can write "reflexive property of equality," like this.0366

But if you are going to use it for congruence of segments, then you are going to have to write "reflexive property of congruent segments"--this or this.0372

And the same thing for the symmetric property; instead of "symmetric property of equality,"0385

like this, using the equals sign, you are going to write "symmetric property of congruent segments."0390

Or you can write out "segments," too.0399

For the transitive property, it is "transitive property of congruent segments," also.0403

This is the theorem; the theorem says that congruence of segments is reflexive, symmetric, and transitive.0408

The theorem is saying that you can use these three properties for the congruence of segments.0414

So, before we can actually use this and this and the transitive property of congruent segments, we have to prove this theorem.0420

Remember, theorems have to be proved; so before we can actually use this, we have to prove this theorem.0429

This is the proof to prove this theorem; and there are three of them--there is reflexive, symmetric, and transitive.0437

But we are only going to look at the symmetric one.0448

For the other ones, your book should show you some kind of proof to the theorem.0454

Or if you want to look at any online things, you can just look that up for the theorem to prove the other two.0460

For this one, the congruence of segments is symmetric; that is what we are trying to prove.0472

We are proving one part of this theorem: AB is congruent to CD, and we want to prove that CD is congruent to AB.0477

So again, this is to prove the congruence of segments; see how this little symbol right here means "congruent."0488

And these are all segments: AB and CD, both of these.0497

We are trying to show that it can be symmetric; that means--see how this shows the symmetric property?0502

Congruent segments can also use the symmetric property; that is what you are proving.0510

The first statement: AB is congruent to CD: the reason for that is "Given"; the first step is always "Given."0516

The next step, AB = CD: from here to here, all that has changed from this to this is that congruence went to equality.0526

This is actually that, whenever you go from equal to congruent, or from congruent to equal, this is "definition of congruent"...0540

and since we are dealing with segments..."segments."0549

If you use angles, then you would say "definition of congruent angles."0553

And then, here, from here to here, what happened? It just flipped.0559

And we know that we can use (since we already used it for the last lesson)--this is the symmetric property of equality.0567

We know that we can use the equality one; so we are going to use that for the reason.0577

And then, the next step: see how it just went from equal to congruent now?0581

Then again, this is "definition of congruent segments."0588

And this is what we are trying to prove: that we can use the symmetric property for congruent segments.0595

Now that we have proven this theorem, we can now use these statements as our reasons.0606

Here is an example of a proof: Given is AB congruent to CD.0618

Now, you use your diagram; if you are given this, and you don't have a diagram, then draw in a diagram.0629

And then, this little mark right here, this hash mark, is just to show that this is congruent.0637

So, if you mark this once, and you mark this once, then they are congruent.0642

The next two that you want to show congruent: BD is congruent to DE.0649

I can't mark it once; if I mark it once, then that is saying that this and this are the same.0656

So then, you have to mark this twice; so this is a different one...and then DE...you mark this twice.0662

So, the segments with the same number of hash marks are congruent.0668

And you are proving that AD, this whole thing, is congruent to CE, that whole thing.0675

We know that their parts are congruent: AB is congruent to CD, that part; and then, this small part is congruent to the other small part.0682

And we want to prove that the whole thing, AD, is congruent to CE.0690

Now, remember: if you are given parts, and you have to try to prove something with the whole thing,0697

any time that you have to go from parts to whole or whole to parts, remember,0706

you are going to use the Segment Addition Postulate.0711

So, if you are dealing with segments (which we are now), then it is going to be the Segment Addition Postulate.0716

If these are angles, and you go from partial angles to the whole angle, or the whole angle to its parts, then you are going to use the Angle Addition Postulate.0722

Just so you are aware of your steps...the first step I have to fill in, and the reason is "Given."0734

Remember: I am only writing this down for my given statements.0748

AB is congruent to CD, and BD is congruent to DE.0753

Now, step 2...and I like to, for my statements and reasons, keep the same statement and the same reason on a level, so that it is easy to see.0769

From step 2's statement to step 2's reason...number 2: AB = CD, and BD = DE.0784

How did I come up with this? Well, all that has changed is going from "congruent" to "equal."0796

AB = CD right here, and BD = DE.0807

So, we just went from congruence to equality; and that means that it is "definition of congruent segments."0811

You can also say "definition of congruency"; sometimes that is used, but for this, we are just going to use "definition of congruent segments."0822

Then, this next step: here, since BD is equal to DE, this was added to AB, and this is added to this.0833

AB + DE = CD + DE; so it is just added onto each of those steps.0856

If you want, if this step is a little bit confusing, you can go on to the next step first,0873

because the next step, right here, uses the Segment Addition Postulate, AB + BD = AD, and CD + DE = CE.0880

We know that step 4 is "Segment Addition Postulate"; and then, from there, you can move on to step 3.0898

Now, proofs can be a little bit different; remember how, last lesson, I gave you the analogy of driving the car from point A to point B--0910

or maybe not driving--maybe you can't drive; but if you are going from point A to point B, you have a series of steps to get there.0920

You can make a left here, and make a right here, and so on.0932

Now, usually, from point A to point B, there might be a couple of different ways that you can get there.0936

So, in the same way, with proofs, it is not always going to be the same.0943

For the most part, there is a best way to get from point A to point B.0950

But you might have a few more steps, and that just means that you took a longer route to get to point B, your destination.0955

So, it doesn't mean that it is incorrect; as long as you correctly get from point A to point B, you have a correct proof.0965

For this one, again, you can just write, or you can just have this before,0980

because that is going to help you move on to the next step, which is AB + DE, and then this.0987

So then, 3 and 4 can be switched around; but you can also think of it as AB, and then you are adding on this BD.0996

And then, you are just taking the same equation right here, and you are just including BD and DE into it.1004

You are just including it in; now, for this one, you can do this in two steps, too.1013

You can say, since BD = DE, to add BD here and add BD here, because then it would be the Addition Postulate.1023

Here, we can also say that this is the Addition Postulate, because, since they are the same, we are just adding it onto this whole equation.1040

Let's just call it the...I'm sorry, not the addition postulate; the addition property of equality.1048

And then, this right here is the Segment Addition Postulate.1060

Then, since we know that all of this equals all of this--look here, I have all of that and all of that.1067

If this equals this, then that is the same thing as this equaling this, which means that AD is going to equal CE.1079

So, my next step is going to be AD = CE, and that is the substitution property of equality, because it is just substituting all of this from step 3.1092

If step 3 is my equation, all I did was replace this part, the whole thing, with AD, and this whole thing with CE; so it is just the substitution property.1113

And then, step 6: my step 6 has to be my "prove" statement.1123

From here to here, it went from equal to congruent; so "definition of congruent segments."1131

And we are done with this proof; so then, from that, we just were able to prove that AD, the whole thing, is congruent to CE.1144

OK, setting up proofs: I want you to practice setting up a proof.1155

You don't have to actually solve the proofs yet; don't worry about the proof itself; we will start with just setting them up first.1160

If you are given a statement or a conditional (an if/then statement), then it is just setting it up, and then having a diagram--having a drawing.1171

This first one: If two segments have equal measures, then the segments are congruent.1187

If you have two segments (let's say AB is one segment, and CD is another segment), and they are the same1196

(let's say this is 10 and this is 10), then the segments are congruent.1209

So, the "if" part, my hypothesis, is going to be my given; that is what is given to me, "if two segments have equal measures."1214

My given is going to be that...now, I am not writing it in words; I am going to write it using my actual segments that I drew out;1225

the given is that two segments have equal measures, and my two segments are AB = (because it is equal to) CD.1237

That is my given: two segments having equal measures.1249

Then, the segments are congruent--I am trying to prove that the segments are congruent.1255

This is my given statement; this is my "prove" statement, because this is the conclusion.1264

So then, the conclusion is the "prove" statement; both segments are congruent, so AB is congruent to CD.1273

When you write segments, only when you are dealing with congruent segments do you write the segment up here.1285

If AB = CD, then you are proving that AB is congruent to CD.1295

And that is all you need to do for now--just setting up the proof in this way: the given statement, the "prove" statement, and then the diagram.1299

Write the given and "prove" statements again, and draw the diagram.1311

Vertical angles are congruent: now, this one is not a conditional statement--it is not written in if/then form.1314

You can rewrite it if you want; or if you are able to figure out1323

what the hypothesis and the conclusion are by looking at this, then you can just go ahead and set up.1327

Vertical angles are congruent: if I draw my diagram, I have vertical angles--let's say 1 and 2.1333

If I rewrite this as a conditional, it is "If angles are vertical," or you can say "if two angles are vertical angles, then they are congruent."1349

My given is that angles are vertical; now, I am not writing this out; I have to write it out using my diagram.1377

That means...well, in this case, I can say that angle 1 and angle 2 are vertical angles; I am using my actual angles.1388

And then, the "prove" statement is that they are congruent; that means that I have to say that angle 1 is congruent to angle 2.1404

Now, you might not use angles 1 and 2; maybe you can label these points, and then use angles that way.1420

You can use whatever angles you want; you are setting it up the way you want to set it up.1427

But just make sure: it has to be written like this; you can just label the angles differently.1431

The next one: If a segment joins the midpoints of two sides of a triangle, then its measure is half the measure of the third side.1440

We have a triangle; if a segment joins the midpoints of two sides...so then, I need midpoints.1450

A segment joins the midpoints: here is one midpoint; that is a midpoint; here is another midpoint.1470

If a segment joins those two midpoints together, that is the segment that this is referring to.1480

Then, its measure, the measure of this (let me just label this: A, B, C, D, E), DE, is half the measure of the third side.1490

So then, these were the first two sides that it was talking about; the third side would be BC.1506

They are saying that, if this segment joins the midpoint of this and the midpoint of this,1511

from the triangle, then the measure of that segment is half the measure of the third side, this unmentioned one.1518

My given is a few different things; my given can be that...1529

Well, first of all, I have a triangle; I need to mention that, so a given is my triangle ABC; that is the triangle.1536

Then, I say that D (because I need to mention its midpoints) is the midpoint of AB, and E is the midpoint of AC.1544

And then, my "prove" statement will be that its measure is half the measure of the third side.1576

So then, DE is half the measure of BC.1583

So, given my triangle ABC, where D is the midpoint of AB and E is the midpoint of AC, then I am proving that DE is half the measure of BC.1598

DE is half the measure of ("of" means "times") BC.1610

If you divide BC by 2, then that is going to be DE.1617

OK, let's do a few more examples: for these examples, you are going to justify each statement with a property from algebra or congruent segments.1623

The first one: if 2AC = 2BD, then AC = BD.1634

This one--what did you do here? You just divided the 2 to get AC and BD, so this one is the division property.1643

Now, it is the division property of equality--we can just write the equals sign.1652

The next one: XY is congruent to XY; remember that this is a theorem that we went over at the beginning of the lesson, using congruency.1661

For this, this one, if you remember, is the reflexive property, but it is not of equality,1674

because it is not an equals sign; it is the reflexive property of congruent segments.1688

The next one: If PQ is congruent to RS, and RS is congruent to UV, then PQ is congruent to UV.1700

See how these are the same, so PQ is congruent to UV; this is the transitive property.1710

Is it of equality? No, it is the transitive property of congruent segments.1720

The next example: Write the given statement, the "prove" statement, and the drawing to set up a proof.1731

Acute angles have measures less than 90 degrees.1745

This is an acute angle; my given statement is going to be something with being acute, so I am going to say that angle A is an acute angle.1759

Remember from this: you are drawing a diagram and labeling it, and you are going to use what you labeled in these statements.1783

The given is that angle A is an acute angle, and then, what you have to prove is that this angle measures less than 90 degrees.1792

To write this as a conditional, you can say, "If the angles are acute, then they measure less than 90 degrees," and that can be our conditional.1803

Angle A is an acute angle, and then you are proving that it measures less than 90 degrees.1837

So, you are going to say, "The measure of angle A is less than 90 degrees."1842

The next example: If two segments are perpendicular, then they form 4 right angles.1856

If I have segments AB and CD, then you are saying that if they are perpendicular, and that means that all four are right angles.1871

We know that I can write that this is a right angle, because they are perpendicular.1897

But remember how this is a linear pair; so this angle and this angle are supplementary, because linear pairs are always supplementary.1902

That is how you would prove this; but again, we don't have to prove anything right now.1918

We are just going to write what is given and what we have to prove.1922

The given statement is that these are perpendicular; AB is perpendicular to CD.1927

And then, prove that (I'll label this: 1, 2, 3, 4) angle 1, angle 2, angle 3, and angle 4 are right angles.1943

Given that AB is perpendicular to CD, because segments are perpendicular, you are proving that they form four right angles.1976

So, angles 1, 2, 3, and 4 are all right angles.1985

We are going to actually do the proof for the last example.1993

Write the two-column proof: Now, you can just write statements and make two columns for the statement and the reason;1999

or you can just draw out your actual two-column proof.2011

Then, let's look at this: GR is congruent to IL; GR, this whole thing, is congruent to this whole thing.2019

SR, this short one, is congruent to SL; I have to prove that GS is congruent to IS.2037

Now, I am not going to mark that yet, because I have to prove it.2050

It is not true yet, until I am done with my proof.2053

My statements...let's do it like that, and then here are my statements; here are my reasons.2061

Step 1 (I'll use a different color): GR is congruent to IL, and SR is congruent to SL; the reason is that it is given.2078

Step 2: Now, look at this again; whenever you deal with whole segments with its parts, we are going to use the Segment Addition Postulate.2102

You want to try to break it up into its parts.2116

GR = GS + SR, and IL = (I am just doing this to both segments) IS + SL; this is the Segment Addition Postulate.2120

Now, before I go on, notice how in step 1, I have congruent signs, and in step 2, I have equals signs.2167

For the Segment Addition Postulate, I have to make it equal, not congruent.2176

If I want to go on any further, using GR and IL or anything else, I need to change my step 1 to equals signs.2181

I could have done that for step 2, and then for step 3 done the Segment Addition Postulate;2194

but as long as you get it done before you use it, it is fine.2198

So then, in my step 3, I am going to write GR = IL, and SR = SL.2204

And then, since all I did was change it from congruent to equal, it is "definition of congruent segments."2215

Then, my next step: See how GR is right there, and then IL is right there;2228

so then, I can use this as my equation, and then substitute in the parts of GR, all of this, because that equals GR.2240

I can substitute all of that into GR, and then substitute all of this for IL.2253

I am going to make GS + SR equal to IS + SL; again, all we did was substitute in all of this stuff for GR, and substitute in all of this for IL.2261

And that is the substitution...of congruent segments?--no, of equality, because it is equal.2283

Then my step 5: Since I know that...2296

Well, from here, let's go back to our previous statement: GS is congruent to IS.2304

Here is GS; here is IS; somehow, I have to get rid of SR and SL.2309

Luckily, I know that they equal each other; so I can substitute in one for the other.2317

GS +...it doesn't matter which one; I can change this to SL, or I can change this one to SR; so let me just change this to SL...equals IS + SL.2324

Step 5 (I am running out of room here): I use the substitution property.2341

Remember not to write "sub." for substitution, because that can be the subtraction property, too.2348

And then, step 6: For me to get rid of SL and SL, I just have to subtract it.2355

If I subtract, then I just get GS = IS; and that is the subtraction property of equality.2366

Now, I am almost done; right here, even though I have GS = IS, and this is GS converting to IS, I have to make it look exactly the same.2379

My last step that I am going to do is GS is congruent to IS; and that step is the definition of congruent segments.2390

And that is it; we have proved that GS is congruent to IS.2412

Remember: we used the Segment Addition Postulate, because I have this whole segment,2418

and then I need to use these parts; for parts to the whole, use the Segment Addition Postulate.2422

If you want, you can try to erase it or try not to look at this, and try working on this proof yourself.2434

Just go back and rewind it or whatever, and then just try to work on the proof on your own.2444

And then, you can come back to this and check your answers.2451

OK, well, that is it for this lesson; we will see you for the next lesson.2455

Thank you for watching Educator.com.2461

Welcome back to Educator.com.0000

This next lesson is on proving angle relationships.0002

Let's go over the supplement theorem; this lesson actually uses a lot of theorems.0007

And remember, from the last lesson: a theorem is a statement that has to be proved.0012

It is not like a postulate, where we can just assume them to be true; the theorems have to be proved in order for us to use them.0020

And usually, your book will prove it for you; and then, after that, you can use it whenever you need to.0029

The first one (this is the supplement theorem) is "If two angles form a linear pair, then they are supplementary angles."0035

Now, supplementary angles are two angles that add up to 180 degrees.0045

A linear pair would be two angles, a pair of angles, that form a line--"linear" means line.0059

Two angles, a pair of angles, that form a line are supplementary angles.0070

That means that if the two angles form a line, then they will add up to 180.0078

If angle 1 and angle 2 form a linear pair (here is angle 1, and here is angle 2)--because you put them together, they form a line--0085

and the measure of angle 1 is 85 (this is 85), then find the measure of angle 2.0095

Then, this is what we are trying to find.0102

Well, we know that, since these two angles form a linear pair, they are supplementary; that means that they add up to 180.0105

And then, if the measure of angle 1 is 85, well, this plus this one is 180, so I can say 85 degrees, plus the measure of angle 2, equals 180.0115

This is the measure of angle 1 that is given; and then, the measure of angle 2...together, they add up to 180.0130

If I subtract 85, what will I get here? The measure of angle 2 is 95 degrees.0138

That is how I am going to find the measure of angle 2--using supplementary angles and the supplement theorem.0150

OK, congruence of angles is reflexive, symmetric, and transitive.0159

Reflexive: remember, reflexive is when a = a; that is the reflexive property.0167

The symmetric property is just "if a = b, then b = a."0173

And then, the transitive property is, "if a = b, and b = c, then a = c."0178

So, if you need to review those properties, then just go back a few lessons before when we talked about each of those in more detail.0185

But congruence of angles: when we talk about angle congruence, then these properties also apply.0197

This is a proof, because this is a theorem, so we have to prove that.0208

And we are just going to prove one of these; so in your book, they will actually show you all of the proofs for each of these,0216

but for this lesson, we are just going to just prove one of them.0224

Let's see, angle 1 is congruent to angle 2: we know that these two are congruent;0229

angle 2 is congruent to angle 3, so prove that angle 1 is congruent to angle 3.0236

Which property does that sound like? It sounds like the transitive property.0243

We are just saying that we are going to prove that the transitive property can be applied to congruent angles.0251

#1: For statements, and then reasons on this side...statement #1: Angle 1 is congruent to angle 2, and angle 2 is congruent to angle 3.0260

The reason for that, we know, is "Given"; the first one is always given.0285

#2: Since we are trying to prove that the transitive property works for congruent angles, I have to first,0291

because I know that the transitive property will work for angles that are equal, change this to measure of angle 1 = measure of angle 2,0304

and measure of angle 2 = measure of angle 3; and in that case, all I did was change the congruence to the "equals."0317

That is the definition of congruent angles.0327

And then, the third one: If the measure of angle 1 equals the measure of angle 2, and the measure of angle 20335

equals the measure of angle 3, then I know that I can apply the transitive property to this one.0341

So, this is "the measure of angle 1 equals the measure of angle 3," and this would be the transitive property.0347

And I know that that is the transitive property.0359

I can use the transitive property, because it is the "equal."0363

And then, #4: I am trying to prove that angle 1 is congruent to angle 3; angle 1 is congruent to angle 3, and all I did there was to go from equal to congruent.0367

So, it is the definition of congruent angles.0381

So then, here, this is the proof to show that if you have this given to you, you can use the transitive property.0388

You can prove that the transitive property works for the congruence of angles, and that is just all it is.0401

Now, you can use these three properties on the congruence of angles.0406

Angle theorems: #1: Angles supplementary to the same angle or to congruent angles are congruent.0416

Now, these theorems in your book are probably labeled Theorem 2-Something; make sure you don't use that name.0426

Don't call it by whatever your book calls it; the only time that you can use the same name is when there is an actual name for the theorem.0435

But don't call it by what the book labels it, if it is 2-Something, because all of the books will be different.0444

And if there is no name for it (like number 1--there is no name for it; it is just the theorem itself), then you would have to write out the whole theorem.0453

But there is a way for you to abbreviate it: Angles supplementary to the same angle, or to congruent angles, are congruent:0462

so then, you can just say, "Angles supplementary to same angle or to congruent angles are congruent."0470

You can just write it like that; any time you have "angles," you just write the angle sign; if you have "congruent," then write the congruent sign.0482

And then, you can just shorten "supplementary."0491

Now, what is this saying? "Angles supplementary to the same angle or to congruent angles are congruent."0496

Well, if I have an angle, say angle 1; and let's say this angle is supplementary to angle A (this is angle A);0504

angle 1 and angle A are supplementary angles--that means that, if you add them up0526

(you add the measure of this angle and add the measure of that angle), you get 180.0532

Now, let's say I have another angle, angle 2: now, this angle is also supplementary to angle A.0537

Then, they are saying that these two angles have to be congruent.0549

If two angles are supplementary to the same angle, then they have to be congruent.0554

And if this is 100 degrees (the measure of that is 100), then this has to be 80.0560

There is only one angle supplement to this, and there is only one angle supplement to that.0569

So, if these two are supplementary, and these two are supplementary, the only angle measure that this can be is 100.0578

I can't find another angle measure that will be supplementary to this angle that is different than 100.0588

As long as they are both supplementary to the same angle, then they have to be congruent.0598

You can't have two angles supplementary with different measures.0602

That is what this is saying: if angles are supplementary to the same angle (both of these are supplementary to the same angle, A), then they are congruent.0611

And the same thing here: Angles complementary to the same angle or to congruent angles are congruent.0626

If I have...here is angle 1, and that is complementary to angle A; and then I have angle 2;0631

if this is, let's say, 70, and angle 1 and angle A are supplementary, then this has to be 20 degrees, because "complementary" is 90.0649

So, if the measure of angle 1 is 70, and the measure of angle A is 20,0670

and let's say that the measure of angle 2 is also complementary to angle A;0675

then this has to also be 70, so they are going to be congruent; angle 1 is congruent to angle 2.0679

More theorems: All right angles are congruent.0694

If I have a right angle, the measure of that angle is 90; if I have a right angle like this, guess what that is--90.0701

If I have one like this, that is still 90; so all right angles are congruent, because they all have the same measure.0718

Vertical angles are congruent: this one is actually going to be used a lot.0730

Remember: vertical angles are angles like this one and this one.0737

These are vertical angles, and these are vertical angles; so that means that these two, the angles that are opposite each other, are going to be congruent.0746

And then, these two angles that are opposite each other are also congruent.0756

Vertical angles are always congruent; they are not congruent to each other--make sure that you don't get it confused with this one and this one.0762

It is always going to be the opposite, so this and this.0772

Perpendicular lines intersect to form four right angles.0778

If I have perpendicular lines, this is 90; this one right here (remember, linear pairs are always supplementary)--0782

if this is 90, then this one has to also be 90, because they form a line.0799

And then, this angle and this angle form a line; that is a linear pair, so this has to be 90; and the same thing here--this has to be 90.0805

They all become right angles; as long as the lines are perpendicular, you have four right angles.0815

Complete each statement with "always," "sometimes," or "never."0827

Two angles that are complementary to the same angle are ___ congruent.0831

Two angles that are complementary to the same angle--I have two angles, and they are complementary to the same angle, A.0839

This is complementary to A, and angle 2 is complementary to A; then they have to always be congruent.0856

Vertical angles are [always/sometimes/never] complementary.0869

Vertical angles, we know, are like this; so, vertical angles are complementary...0873

well, we know that vertical angles are congruent; that means that,0884

if they are to be complementary, then that means they have to add up to 90.0887

But then, they have to be the same measure; so if they are going to add up to be 90, then each of them has to be 45.0894

In that case, then the vertical angles would be complementary, because they add up to be 90 degrees.0904

But look at these angles right here: these angles are not complementary, so what would this angle be?0911

This would be 135; how do I know that?--because this is a linear pair, and they are supplementary.0921

This one and this one...135 + 45 has to add up to 180, because it is a line; they form a linear pair.0930

This angle and this angle are congruent, because they are vertical; vertical angles are always congruent.0939

But they are not complementary; so in this case they can be complementary, but in this case they are not.0948

So, here, this would be my counter-example--an example that shows that something is not true,0955

an example that shows where this statement is not going to be true.0963

Then, I know that it could be complementary, and it might not be complementary.0969

So, this will be sometimes.0976

Two right angles are ___ supplementary; two right angles are [always/sometimes/never] supplementary.0982

If I have a right angle, that is 90; if I have another right angle, that is 90; 90 and 90 always make 180.0992

So, this will be always; two right angles are always supplementary.1004

OK, let's do a few examples: Two angles that are supplementary [always/sometimes/never] form a linear pair.1015

Let me think of a counter-example: try to think of two angles that are supplementary that do not form a linear pair.1028

Well, how about if I have an angle like this; let's say this is 100 degrees, and then I have another angle like this that is 80 degrees.1039

They are supplementary, right? Yes, they are, because they add up to 180.1057

Do they form a linear pair? No, so this is my counter-example.1061

Can they form a linear pair, though?--yes, because, if I have a linear pair, here is 100 and here is 80.1075

So, my answer will be sometimes.1088

Two angles that form a linear pair are [always/sometimes/never] supplementary.1097

Two angles that form a linear pair--that means that we have to have a linear pair--are [always/sometimes/never] supplementary.1102

It is always, because a linear pair will always be supplementary.1116

I can make this 120; then this will be 60; if I make this 100, it is going to be 80.1123

No matter what, they have to be supplementary, because they form a linear pair.1129

Find the measure of each numbered angle: the measure of angle 1 is 2x - 5; the measure of angle 2 is x - 4.1137

What do I know about the measure of angle 1 and the measure of angle 2?1147

They form a linear pair, so then, they are supplementary; they add up to 180.1150

Make sure you don't make them equal to each other, because they are not.1159

This is obviously an obtuse angle, and this is an acute angle.1164

And they don't look like they are the same; besides that, you just know that you can't always assume1169

that two angles that are adjacent are going to be congruent, or that they are going to be equal or have the same measure.1175

Keep in mind that two angles that form a linear pair--always remember that a linear pair's angles are going to add up to 180.1184

So, I am going to take the measure of angle 1, 2x - 5; I am going to add it to the measure of angle 2: + x - 4.1192

And then, I am going to make them equal to 180.1201

See how this was the measure of angle 1 and the measure of angle 2, like that.1205

And then, I just solve it out; so then, 2x + x is 3x; -5 - 4 is -9; that equals 180.1218

I add the 9; it becomes 189; and then I divide by 3, and that is going to give me 63.1229

OK, so then, that is my x; and then, what is it asking for?1242

They are asking for the measure of each numbered angle; so then, be careful--when you solve for x, you don't leave it like that.1250

You have to plug it back in; if they were asking for x, then yes, that would be the answer.1259

But in this problem, they are not; they are asking for the measure of the numbered angle.1266

Then, you have to plug it back in, because they want you to find the measure of angle 1 and the measure of angle 2.1271

So then, for the measure of angle 1, 2x - 5...I have to do 2(63) - 5; and that is going to be 126 - 5; that is 121, so it is 121.1277

The measure of angle 2 is 63 - 4, which is 59.1302

And then, make sure; you can double-check your answer by adding them up, because if we add them up, then they should add up to 180, and this does.1309

So, here is the measure of angle 1, and here is the measure of angle 2.1320

Measure of angle 3 and measure of angle 4--their relationship: they are vertical.1329

We know that vertical angles are congruent; so in this case, since they are vertical, I can make them equal to each other.1336

This is not supplementary, and don't assume that they are complementary.1347

They could be supplementary, and they could be complementary, but we don't know that they are.1352

What we do know for sure is that they are congruent; they are the same.1358

So, I can just make them equal to each other.1362

So, the measure of angle 3 is going to equal the measure of angle 4.1366

The measure of angle 3 is 228 - 3x = x; that is the measure of angle 3, and this is the measure of angle 4.1374

If I add 3x to that side, I am going to get 228 = 4x.1386

And then, I divide the 4, I am going to get 57.1391

And then, again, they want you to find the measure of the numbered angle, not x.1402

So, we take the x, which is 57, and we are going to plug it back in.1414

The measure of angle 3 equals 228 - 3(57); and then, 228 minus...this is going to be...171.1419

And then, subtract it, and you are going to get 57; and the measure of angle 4 is going to be 57.1450

Is that right? Well, what do we know?1477

Now, unlike this problem right here, where we can just add them up and then see if they are supplementary,1480

because we know that they are supplementary, we can't do that to this, because they are not supplementary; they are congruent.1484

We have to check our answer; we have to just see that they have the same measure.1492

And they do; so then, that would be the answer for the measure of angle 3 and the measure of angle 4.1496

OK, find the measure of each numbered angle; angle 1 and angle A are complementary; angle 2 and angle A are complementary.1505

Since we know that this is complementary to this and this is complementary to this, what do we know?--1519

that the measure of angle 1 equals the measure of angle 2.1525

If two angles are complementary to the same angle, angle A, then they are congruent.1530

Angle 1 is complementary to angle A; angle 2 is complementary to angle A; that means that these have to be congruent.1538

I can just make these angles equal to each other, and then I just substitute in 2x + 25.1546

And then, this is going to be x = 20.1559

And then again, I have to look back and see: OK, they want me to find the measure of each number angle,1567

so I have to find the measure of angle 1 and the measure of angle 2.1572

I have to plug x back in: so the measure of angle 1 equals 3(20) + 5.1575

Let's see: here I am going to have 65, and then, even though I know that they are the same value,1584

that the measure of angle 1 is going to equal the measure of angle 2, I still want to plug in x and see if I am going to get the same number.1600

2 times 20, plus 25--this is 40, and that is 65.1610

So then, I do have the same measure, so that shows me that it is right.1617

The next example: here, we have a proof; we are going to write a two-column proof.1628

Our given statements are that the measure of angle ABC is equal to the measure of angle DEF, or DFE; this should actually be DEF;1635

and the measure of angle 1 equals the measure of angle 4.1656

We are going to prove that the measure of angle 2 is equal to the measure of angle 3.1661

So, remember: my column here is going to be statements; my column here is going to be reasons.1665

Remember: my first step is going to be the given statements: so the measure of angle ABC equals the measure of angle DEF.1680

And then, the reason for that is going to be "Given."1694

Number 2: Let's look at this really quickly, or think about this.1700

I have that this big angle is equal to this big angle; they have the same measures.1709

And this part, this angle, is equal to the measure of this angle.1717

And then, I want to prove that the other part of it is going to be equal to this part.1727

Remember: when we have a bigger angle, and we want to break it down into its parts1733

(because that is what we are doing: we are dealing with the big angle's parts, and the same thing on this side),1738

then we want to use the Angle Addition Postulate, because that is what breaks it down from its whole to its parts.1746

The measure of angle ABC equals the measure of angle 1, plus the measure of angle 2.1758

The measure of angle 1 plus the measure of angle 2 equals this big thing.1770

And the same thing for the other one: the measure of angle DEF equals the measure of angle 4, plus the measure of angle 3.1773

And the reason is the Angle Addition Postulate.1797

Now, from here, since I know that the measure of angle 1 is equal to the measure of angle 4,1810

and I want to prove that these two are equal to each other, I can just make this whole thing equal to this whole thing.1815

So, I am going to use step 1, and I am going to replace all of this with its parts.1825

This is the formula; this is equal to this--that is what is going on with this one: the measure of angle ABC is equal to the measure of angle DEF.1835

I am going to replace the measure of angle ABC with its parts, which is this,1846

and then replace all of that with its parts there.1854

And the reason is going to be the substitution property--and I can write "equality."1862

And then, again, the measure of angle 1 and the measure of angle 4 are equal to each other.1876

So, I can just replace one of them with the other; so I just replace this with the measure of angle 4,1884

because it is given; and I should actually write that, too, for here, because this is another given statement:1898

the measure of angle 1 equals the measure of angle 4.1906

Since they are equal to each other, I can just replace one for the other.1911

And whenever I do that, that is also the substitution property (I am running out of room).1917

Then, here, since these two are the same, I can just subtract it out.1925

If I subtract it out, then, the measure of angle 2 equals the measure of angle 3.1934

And right here, this is the subtraction property of equality.1941

And is that my "prove" statement? Yes, it is, so I am done.1956

So, remember to always keep in mind...because, from here, you can go in a lot of different directions;1962

you can make a left; you can make a right; you can take any direction; it depends on your destination--where are you trying to get to?1970

You are trying to get to this right here; if you are trying to get to this statement right here, you have to lead it from this step to this step.1979

Maybe after you write your given statement, or before you even start, just look at it and think,1991

"OK, well, if I want to get from here to here, what steps do I have to take--what do I have to do?"1997

And once your last statement is the same as this right here, then you are done.2005

That is it for this lesson; thank you for watching Educator.com--we will see you next time.2013

Welcome back to Educator.com.0000

This next lesson is on parallel lines and transversals.0002

Let's go over some lines: first of all, parallel lines are two lines in a plane that never, ever meet.0008

We know that, if we have arrows at the ends of these lines, then that means that it is going on forever, continuously.0024

And no matter how long these lines are, no matter how far out they go, they will never intersect; they will never meet.0030

They are parallel; and the symbol for parallel would be two lines like this.0039

So, if I want to say that line AB is parallel to line CD, then I can say that line AB is parallel to CD; and that is how you can say it.0046

So then, this symbol right here is for parallel.0056

Also, when you have two lines drawn like that, to show that they are parallel, you can draw little arrows like that on the lines.0063

And that shows that these two lines are parallel.0074

If I want to draw two other lines that are parallel to each other, but not parallel to these,0076

then instead of drawing them one time (because, if I draw it one time, that means that all three of these would be parallel),0084

only the ones that I have drawn once are parallel together; and then, this one I have to draw twice.0091

And that shows that if I drew two for this one and two for this one, then these two are parallel.0098

If these two are not parallel, then you could just either not draw them or, to show that they are not parallel,0105

this could be drawn twice, and this you can draw three times.0112

And then, you know that those would not be parallel, because it has two, and this one has three.0116

The next one: skew lines are two lines not in the same plane that will never intersect.0123

So, they are like parallel lines in that they don't intersect; that is the only thing that they have in common.0131

But they are not in the same plane.0136

If you hold two pencils like this--you have a pencil like this, and you have a pencil like this--look: they are not in the same plane, and they do not intersect.0142

So, these are skew lines--two lines that would never intersect, but they are not parallel.0152

They are not parallel; they do not intersect, because they are not in the same plane.0160

I can also draw this; if I draw a box, then I can say (let me draw this) that maybe this side and this side are skew, because they are not in the same plane.0166

There is no way that I can draw a single plane that will contain those two lines.0193

So then, those would be skew lines; another pair of skew lines would be maybe this one right here, the back one, and this one.0204

These two will never intersect, and they are not in the same plane.0213

Skew lines that are not in the same plane do not intersect, and they are not parallel.0216

Transversal: a transversal is a line that intersects two or more other lines in a plane.0224

It doesn't matter if these two lines are parallel or not.0236

Let's say that these two lines are not parallel; they are just two lines.0241

And a line that intersects both of them, two or more, is a transversal; so the red line is the transversal.0245

And that is...we are going to be using transversals in this lesson.0261

Here are two lines, and they could be parallel; they don't have to be parallel, so don't assume that they are parallel.0272

Even if they are drawn side-by-side, even if they look parallel, do not assume that they are parallel unless they tell you.0280

And then, this line right here would be the transversal, because it is the line that is intersecting two or more lines.0287

When a transversal intersects two or more lines, there are these angles that are formed.0298

Now, the angles, when we have a pair of them--they form relationships.0306

And these relationships have names; special angles have special names.0313

Now, within the two lines that the transversal is intersecting, we look at the two lines0319

(that is this one and this one), and if I label these lines and say that this is l, p, and let's say q,0326

and lines l and p are intersected by the transversal; all of the angles between the two lines l and p0338

are interior angles, because the angles are inside the two lines; they are within the two lines, in between.0346

The interior angles would be angle 3, angle 4, angle 5, and angle 6; they are all interior angles, because they are inside the two lines.0354

Exterior angles are all of the angles that are outside the two lines: angle 1, angle 2, angle 7, and angle 8.0369

Exterior, exterior; 3, 4, 5, and 6 are all interior; and then these two are exterior.0382

Then these four are the actual special names based on the relationships between the angles.0390

We know that numbers 2 and 3 are vertical angles; 2 and 4 are adjacent angles, or they could be a linear pair.0397

They are also supplementary angles (just a review of those relationships there).0409

But when we compare an angle from this part to this part, that is when we have the special names.0415

The first one: consecutive interior angles are going to be..."consecutive" just means that they are right next to each other, so the one right after.0426

And they are interior angles: these two words will mean something--we know that interior angles are angles in between the two lines, l and p.0443

4 and 6 are consecutive (it just means that they are next to each other--this angle 4 and this angle 6 are next to each other).0460

That means they are the angle right after each other; and it is all on the same side of the transversal.0468

So actually, another name for this is "same-side interior angles."0475

Some textbooks will call it "same-side interior angles"; but the main name that it is mostly called is "consecutive interior angles."0490

Angles 4 and 6, and the other angles would be 3 and 5...these have to be paired up.0502

It is a relationship between the two angles; so angles 4 and 6 are consecutive interior angles, and then 3 and 5 would be consecutive interior angles.0522

They have to be consecutive, one right after the other, angle 3 and angle 5.0534

And they are inside the two lines; 3 and 5 is one pair of consecutive interior angles, and angles 4 and 6 are another pair of consecutive interior angles.0541

The next one: alternate exterior angles: "alternate" means that they are alternating sides of the transversal.0554

Again, remember: these are pairs; it is a pair of angles that is called alternate exterior angles.0563

So, if one angle is on the right side of the transversal, then the other angle has to be on the left side of the transversal.0568

So, here is a transversal; we know that this is a transversal.0577

It doesn't matter which one, if it is left/right or right/left; it doesn't matter.0589

But then look: "exterior angles"--that means that it is on the outside of the two lines--not the transversal, but the other two lines.0594

So then, always for these special types of angle relationships, there are always at least three lines involved,0603

the two lines and the transversal that is cutting through both lines--at least three.0613

Alternate exterior angles are like angle 2 and angle 7, because one is on the right side of the transversal,0621

and one is on the left side of the transversal, but then they also have to be on the outside of these two lines.0632

So then, angle 2 is outside and on the right; angle 7 is outside and on the left;0640

so alternate exterior angles would be like angle 2 and (and that is just an and sign) angle 7.0647

Also, another pair would be angle 1 and angle 8.0657

Angle 2 and angle 7, and angle 1 and angle 8: there are two pairs.0670

Now, that is alternate (left/right; right/left) exterior angles.0676

The next one is alternate (again, meaning left/right, right/left of the transversal) interior angles.0683

Interior means that it has to be within the two lines; so if I have, let's say, angle 4 and angle 5, they would be alternate interior angles,0693

because they are alternating; one is on the right, and one is on the left of the transversal;0706

and they are both interior; they are both on the inside of the two lines.0711

So, angle 4 and angle 5...and then the other pair would be angle 3 and angle 6; angle 4 and angle 5, and angle 3 and angle 6.0716

Corresponding angles: now, here we don't have the words "consecutive" or "alternate," and we don't have the words like "interior" or "exterior."0736

Corresponding angles are a little bit special; this one is typically the one that students get a little confused with.0748

But just think of it as the angles on the same side of the intersection.0759

Let's pretend that this right here is an intersection, like a street intersection.0768

And let's say angle 2 is on the top right corner of the intersection; then, from this intersection, the same position would be corresponding angles.0774

So, the top right is angle 2; the top right is angle 6; so angle 2 and angle 6 are corresponding angles.0790

From the two intersections, if this is one intersection and this is another intersection,0799

then it is the two angles that are in the same position, the same location, located in the same top right corner, bottom left--whatever it is.0803

So, for angle 1, what is corresponding with angle 1?0816

Well, angle 1 is in the top left corner of the intersection; angle 5 is in the top left corner of this intersection.0821

So, angle 1 and angle 5 are corresponding angles.0828

Angle 1 and angle 5, and then angle 2 and angle 6, angle 3 and angle 7, and angle 4 and angle 8--those are corresponding angles.0833

Again, we have exterior angles, angles in the outside of the two lines; interior angles, all of the angles inside;0860

consecutive, or same-side, interior angles, would be like 4 and 6--they both have to be on the inside and on the same side;0867

so if one is on the right, the other one has to be on the right; or if it is on the left, then it has to be on the left.0881

4 and 6 are consecutive interior angles; 3 and 5 are consecutive interior angles.0889

Alternate exterior angles, like angle 2 and angle 7, have to be alternating and on the outside; angles 1 and 8 are alternate exterior angles.0895

3 and 6 are alternate interior angles.0908

Corresponding angles are the angles that are located in the same corner of each of the intersections, so angle 2 and angle 6, or angle 3 and angle 7.0912

Those are the special angles formed by a transversal.0924

OK, so then let's name the relationship between each pair of angles.0931

Angle 2 and angle 7 (let's write these out); angle 1 and angle 6 (I am just going to write the angle signs on that)...0935

OK, angles 2 and 7: well, look at these lines really quickly.0950

We have this line; let's label this line p and label this line q, and then we can label this line, say, s.0961

We know that the only line that intersects two or more lines is line s, this line right here.0978

So, that has to be the transversal; these are the two lines that are intersected by the transversal.0984

So, when we look at this, even though it looks a little bit different than the diagram from the previous slide,0991

we know that these four angles are the interior angles, because the two lines are p and q.0998

It is not the transversal, because there is no line next to the transversal that can block in angles and have angles on the outside of them.1006

2, 4, 5, and 7 are all interior angles; 1, 3, 6, and 8 are all the exterior angles.1018

So then, angles 2 and 7--well, we know that they are inside, and they are alternating sides of the transversal.1025

So, this is alternate, and then interior angles.1036

Angle 1 and angle 6: now, they are on the same side; they are both above line s, so they are on the same side.1050

But they are on the outside, so they are on the exterior.1068

Now, there is no relationship that is titled "same-side" or "consecutive exterior angles."1071

There is no relationship like that, so in this case, since we don't have a special name for that, we would just say that they are exterior angles.1081

They are just exterior angles; they are both on the outside.1089

Angle 4 and angle 8: if you look at this, if this is an intersection and this is an intersection, they are both on the same corner.1096

4 is on the bottom right; 8 is on the bottom right; so these are corresponding angles.1108

Angles 1 and 8 are alternating, so alternate exterior angles; they are alternating, and they are on the outside.1124

Angles 2 and 5: they are on the same side, and they are in the interior, so these are consecutive interior angles.1142

Or you can call them same-side interior angles, depending on what your textbook calls it.1158

So, let's go over a few more examples: Describe each as intersecting, parallel, and skew.1167

Intersecting, we know, is when two lines meet; parallel is two lines in a plane that do not meet;1175

skew lines are lines that are not in the same plane, and they do not intersect, and they are not parallel.1182

The first one, railroad tracks: Railroad tracks, we know, go like this.1191

Or you can look at the lines that go this way; either way, these two lines are parallel lines.1201

Floor and wall of a room: If we have a floor, and we have the wall, they are intersecting; they are like intersecting planes.1215

Number 3: The front and side edges of a desk--if I have a desk, the front and the side edges are intersecting--they intersect right here.1241

The flagpole in a park and the street you live on are never going to intersect;1263

the flagpole is inside a park that is going this way, and your street that you live on is going this way, so they are skew.1273

The next example: Draw a diagram for each: Two parallel planes.1298

Now, we talked about parallel lines; planes can also be parallel, as long as they don't ever meet--they don't intersect.1304

I can draw a plane like this, and then maybe a plane like this; they won't ever meet.1315

Two parallel lines with the plane intersecting the lines: we have a pair of parallel lines, and then,1328

if I have a plane (now, I am a horrible draw-er, but let's just say that the plane goes like this,1342

where this meets right here and this meets right here), you can say that the plane is intersecting the lines.1357

And these are parallel, so I am going to show that they are parallel, like that.1371

Two skew lines: now, if I just draw two lines that look skew, since I am drawing on a flat surface,1375

it looks like these are going to intersect, because they look like they are on the same plane.1388

So, to draw skew lines, the best way to draw skew lines would probably be to draw a box, like a cube.1395

And then, I can say that maybe this side right here, and then this side right here, are skew.1414

Or I can say (probably a better picture would be) this side right here and this front side right here, the bottom part.1426

So then, this line and this line are skew lines, because they are not in the same plane, and then they will never intersect.1443

The next example: Name each of the following from the figure.1454

Number 1: Two pairs of intersecting planes: Let's say plane CABD and plane...if this right here is on a plane, I can say plane AEB.1458

Those two are intersecting, so I can say that those two planes are intersecting.1480

Even though planes are usually shown by a four-sided figure, and this is only three, we can just say that this will be on that plane.1486

So, I can name a plane by three of the points that are non-collinear: E, A, and B are three non-collinear points,1497

so I can just label a plane by AEB; so then, this bottom plane right here is ACBD; that and plane AEB can be two planes that are intersecting.1505

Another pair--it could be any of the other ones; they are all intersecting.1518

You can say any of the two sides, unless it is the front and the back.1523

The front and the back are not...actually, they are intersecting right here; so any of the two planes on here works.1531

All pairs of parallel segments: OK, well, we know that parallel segments, or parallel lines1542

(segments are just like lines, but they have two endpoints; they don't go on and on forever and ever;1550

they have an endpoint and stop, so these would be considered segments instead of lines)...1556

now, we know that these segments right here are not parallel, because they intersect; segment AE intersects with segment BE;1567

so, those are not parallel segments; all of these that are going up all intersect together, so they are not parallel.1577

The only pairs of parallel segments would be CD; I can say that CD is parallel to AB; I can also say that AC is parallel to BD.1586

Those are the pairs of parallel lines: these two are never going to intersect, and these two are never going to intersect.1606

This one right here can just be plane ABD, or ACBD, and plane AEB.1616

Number 3: Two pairs of skew lines--so then, skew lines, I know, are two lines that are not on the same plane and do not intersect.1636

I can say that AC and (and don't write this symbol right here, because that means "parallel") BE, right here, are never going to intersect.1649

Or I can say that CD and AE are skew lines.1668

And all intersecting planes are all of the sides--the planes that contain the sides of these.1682

So, I can say plane AEB, plane BED, plane DEC, and plane CEA; those are the four planes.1689

All right, the next one: Identify the special name for each angle formed, and state the transversal that formed the angles.1718

OK, let me just label these lines and say that this is l; this will be k; this can be n, and this will be p.1726

Those are the four lines; that way, when you name the transversal, then you can just name it by the line.1745

So, 6 and 09 here is 6, and here is 15.1756

Well, now, we have four lines here; so first of all, remember: to figure out special relationships between these two angles, we only need three lines.1767

We have four here; so for each of these, you have to be able to identify which lines you are using and which lines you are not.1782

And whatever lines you are not using, ignore it or cover it up so that it doesn't confuse you.1793

If we are looking at angle 6 and angle 15, then I am looking at this line right here,1801

because angle 6 is formed from this line, line n, and line k.1808

Those are the two lines involved; and then, remember, there are 3 lines, angle 15 has line p involved.1814

For the first ones, 6 and 15, we know that n, p, and k are involved.1823

Now, line l is not involved for this pair of angles; so I can cover the whole side up, or just ignore it--1829

pretend that this line, line l, doesn't exist--that it is not there, because that is going to confuse you.1842

We are just going to look at this part right here; and from this part, angle 6 and angle 15, the line that they have in common is k.1848

So, k is the transversal, because it is the line that is intersecting two or more lines, lines n and p.1863

Line k is intersecting lines n and p.1872

OK, line n would be a transversal, but only when dealing with an angle that involves one of these and one of these.1875

Then, line n would be considered the transversal, because that would be the common line between one of those sets of angles.1884

But for this problem, we are looking at angles 6 and 15; so then, line k would be the transversal.1894

For 6 and 15, they are alternating sides; one is on the right, and one is on the left, of k.1903

And they are on the exterior, because 7, 8, 13, and 14 are on the inside; they are all interior.1909

6 and 15 are the exterior angles; so this would be alternate exterior angles.1917

And then, the transversal is going to be k.1930

9 and 13: look at 9, and look at 13; angles 9 and 13 involve line l, line p, and line k, but no line n.1941

Line n is not involved in this one; so we ignore line n.1960

Cover it up if it is confusing; make sure that you don't look at it--nothing for these two angles involves line n.1965

So then, if we are looking at 9 and 13, the transversal is going to be line p.1975

9 and 13 would be corresponding angles, because, remember: they are on the same corner of these intersections.1984

9 is the top left; 13 is the top left; so these are corresponding (I will just write out the whole thing) angles.1994

And then, the transversal would be line p.2005

4 and 5: again, these two angles are not involving line p, because angle 4 is formed from lines l and n;2014

5 is formed from k and n; so p is ignored.2028

So then, the transversal of 4 and 5 would be n, because n is the line that is intersecting both of these lines, the other two lines involved.2034

So, from the other two lines that are not the transversal, the inside would be these four right here: 2, 4, 5, and 7;2045

4 and 5 are alternating, because one is on the bottom side, and 5 is on the top side.2054

And they are on the inside; so it is alternate interior angles.2061

And then, the transversal is going to be n.2072

3 and 9 are right here; again, this line, this line, and this line are involved, but no k.2080

And these are consecutive interior angles, because they are consecutive (or same-side), and they are the inside angles.2093

And the transversal between 3 and 9 is going to be line l, because that is the line that is intersecting both of these lines.2112

1 and 6 involve this line, this line, and line n, with line n being the transversal, because that is the one that intersects both of these lines.2121

And again, these are the same side, but we don't have a name, a special relationship, for same-side or consecutive exterior angles.2134

We don't have that one, so these are just going to be exterior angles--they are just both on the outside.2143

That is all that it is saying: they are exterior angles, and the transversal is n.2150

11 and 14 involve this line, this line, and this line; ignore line n.2159

They are alternate, because one is on the bottom, and one is one the top, switching sides of the transversal, p; and they are on the outside.2169

So, they are alternate exterior angles, and the transversal is p.2181

Again, make sure, when you are looking at special angles that are formed from a transversal,2196

that you are only going to be looking at three lines; if you have more than three lines2202

(sometimes you might have more than 4; here you only have 4, so there is only one line to ignore;2207

other times you might have more)--just make sure that you look at the two angles; look at what lines are involved.2213

What is your transversal? Ignore all of the other lines that are there, because it is just going to confuse you.2222

You can cover it or just really pretend that it is not there; and do that for each of the problems.2228

And that is going to make it a lot easier for you.2236

That is it for this lesson; we are going to go over more theorems and postulates in the next lesson on transversals and these special angles.2240

Thank you for watching Educator.com; I'll see you soon.2253

Welcome back to Educator.com.0000

The next lesson is on angles and parallel lines.0002

OK, last lesson, we learned about the different special angle relationships, when we have a transversal.0007

The transversal with the other lines forms angles, and those pairs of angles have special relationships.0017

And one of them was the corresponding angles.0030

Now, the two lines that the transversal cuts through--remember: I said that the lines can be parallel, but they don't have to be.0040

So, even if the lines are not parallel, you are still going to have corresponding angles.0050

But then, now the postulate is saying that, if the lines are parallel, then the corresponding angles are congruent.0056

If these lines are parallel (let's say that they are parallel lines), then each pair of corresponding angles is congruent--only if the lines are parallel.0068

If we don't have parallel lines--if I have lines like this and like this--they are not parallel;0083

they don't look parallel, but I have a transversal--let's say 1 and 2: these angles are corresponding angles, but they are not congruent.0093

They are not congruent, but they are still corresponding; angles 1 and 2 are corresponding angles, but they are not congruent.0105

They are just called corresponding angles; so be very careful--only if the lines are parallel, then you can see that corresponding angles are congruent.0112

They are the same; they have the same measure; they are congruent.0123

Since these lines are parallel, I can say that angles 1 and 2 are congruent.0127

So, angle 1 is congruent to angle 2; and it goes with all of the pairs of corresponding angles,0137

like this one and this one--they are congruent...this one and this one, and this one and this one.0145

Each of the pairs of corresponding angles is congruent only if the lines are parallel--that is very, very important.0152

And that is a postulate; a postulate, remember, is any statement (such as this) that we can assume to be true.0159

It doesn't have to be proved; if it is a theorem (the next few are actually going to be theorems),0170

then they have to be proved in order for you to be able to use them, because it is not true until it is proven.0176

The next one: here is a theorem; now, we are not going to prove these theorems now, but they are shown in your textbooks.0186

The alternate interior angles theorem--just so you know, some kind of proof has to be shown for the theorems0198

in order for them to be counted as true and correct, and then, that is when we can use them.0208

But for now, since they are proven in your book, we are just going to go ahead and use them.0214

The alternate interior angles theorem says that, if two parallel lines are cut by a transversal0219

(meaning, if the two lines that are cut by a transversal are parallel), then each pair of alternate interior angles is congruent.0225

Again, from the last lesson: if I have two lines...now, I know I am repeating myself a lot,0238

but that is so that you will understand this, because I have seen a lot of students make careless mistakes with these,0245

always thinking that these are congruent; in this case, if I tell you that these lines are not parallel,0258

or if I don't even say anything about them being parallel, then you don't assume that they are parallel.0265

We just have to assume that they are not parallel; then we can't say that angle 1 and angle 2 are congruent.0272

We can say that they are alternate interior angles; that is the relationship; but they are not congruent in this case.0277

So, for lines being parallel (now I am telling you that the lines are parallel), then alternate interior angles0286

(let's say that this is angle 1 and angle 2)...angle 1 is congruent to angle 2.0296

If the lines that are cut by a transversal are parallel, then alternate interior angles are congruent; and that is the theorem, the other one.0306

The next one is the consecutive interior angles theorem: If two lines that are cut by a transversal are parallel,0317

then (this is the tricky part--not tricky, but this is the part that students really make mistakes on) the consecutive interior angles0331

are not congruent; they are supplementary--this is very important.0344

Consecutive interior angles, we know, are angles that are on the same side, like these two angles right here.0351

And they are both on the inside, the interior.0357

So, angles 1 and 2 are consecutive interior angles; but then, they are not congruent--they are supplementary.0362

Only if the lines are parallel, then consecutive interior angles are supplementary.0369

See how the other ones that we just went over are congruent: these are not congruent--they are supplementary.0375

You have to say that the measure of angle 1, plus the measure of angle 2, equals (supplementary means) 180.0383

That means that this angle measure, plus this angle measure, equals 180--very important.0391

And the next one: Alternate exterior angles, if the lines are parallel, are congruent.0404

So, here is a pair of alternate exterior angles; angle 1 is congruent to angle 2.0415

And that also works for this pair of alternate exterior angles, like 3 and 4; those will be congruent, also.0425

Here we have parallel lines that are cut by a transversal.0442

If AB (let's say that this is A, and here is point B--and these are the points, not the angles;0445

here is point C and point D...then AB is a line, so it is line AB) is parallel to line CD,0460

and line CA is parallel to line DB (and then I am going to add these parallel markers;0471

that means that these two are parallel lines, and then for these--this is another pair of parallel lines,0484

so that means that I have to draw two of them for these, because it is another pair), find the values of x and y.0488

So then, here we have 80; and then I need to take a look at x.0500

If I look for a relationship between this one and another one, even though these two have a relationship,0508

this has a variable x, and this has a variable z.0517

I would rather use this relationship, 4x and 80, because, if I am going to compare them, at least this one doesn't have another variable.0521

So, it is easier to solve; so then, if I look at these two, I am only dealing with this line, line AB, line CD, and line BD.0530

That means that line AC, I am going to ignore, because it is not involved in this pair of relationships.0543

Remember from the last lesson: you look at the pair, and when you have the special pair, it only has three lines involved.0549

It only has line AB, line CD, and line BD involved; the other lines that are there--cover them up.0560

Those lines are there for another pair of relationships, so just cover it up.0567

You don't need this line for this pair, so just ignore it.0573

And then, to solve it, the theorem (and the relationship between these two: they are alternate interior angles,0583

because BD would be the transversal between these two lines) says that if the two lines0596

cut by a transversal are parallel (which they are--we know that because it gives us that in here),0604

if the lines are parallel, then alternate interior angles are congruent.0611

Since the lines are parallel, I can say that these two angles are congruent.0618

Then, they are congruent, so 4x = 80; and I divide by 4: x = 20.0621

There is my x-value; and then, for my y, let's look at this one.0638

Well, with this one, I know that, since I have an 80 here, 80 is also congruent to this angle right here,0651

because they are corresponding, and I know that these two lines are parallel.0667

If these two lines are parallel, here is my transversal; that means that this angle right here and this angle right here are corresponding.0670

And as long as the two lines that are cut by the transversal are parallel, then corresponding angles are congruent.0680

So then, I can just write an 80 in here; and then, between this and this, they are vertical.0685

Now, I could have just done this angle right here to this right here; so there are many ways to look at it.0694

You can look at corresponding angles; if you didn't really see the alternate exterior angles--0700

if that is kind of hard for you to see--then you can just say that, OK, they are corresponding, and then these two are vertical.0707

And vertical angles, remember, are always congruent.0711

So, you can say that these two are the same, because they are vertical.0717

Or you can say that this and this are the same, because they are alternate exterior angles.0722

And those are the same, as long as the two lines are parallel.0727

So, either way: 4y + 10 = 80; then 4y = 70; so y = 35/2.0730

And that is just 70/4, and then you just simplify it to 35/2.0755

Now, it doesn't ask for the value of z, but let's just go ahead and solve it.0765

We know that 6z and 80 have a relationship.0774

Now, I know that this is 80, because we found x; x is 20; and 4 times x is 80;0783

and also because they are alternate interior angles, so whatever this is, this has to have the same measure.0790

So, either way we look at it: we can look at it as 6z with this one right here,0798

or we can look at this one with this one right here--same relationship, same value,0801

which also means that this one is also the same as 4x; this angle and this angle have the same measure.0806

Either way, the 6z with this angle right here are consecutive interior angles, or same-side interior angles.0814

Now, if the lines are parallel (which they are), then consecutive interior angles are supplementary--not congruent, but supplementary,0826

which means that I can't make them equal to each other.0839

Consecutive interior angles are the only ones that are not congruent from the special pairs of angles.0844

Supplementary--that means that I have to make 6z + 80 equal to 180.0851

6z = 100; z = 100/6, and then I can just simplify this to 50/3, and that is it; that is z.0859

OK, the last theorem from this section in this lesson is the perpendicular transversal theorem.0897

Perpendicular, we know, are two lines that intersect to form a right angle.0904

So, if I have a line like this and a line like this, and they form a right angle, then they are perpendicular.0911

But then, here we have a transversal involved; so in a plane, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other.0918

Here are my parallel lines; I am going to show it by doing that.0932

If my transversal, which is this line right here, is perpendicular to just one of the lines0936

(it doesn't matter which one), as long as these lines are parallel (they have to be parallel),0943

if it is perpendicular to one of the lines, then it has to be perpendicular to the other line.0951

If this is perpendicular to this line, then it is going to be perpendicular to this line, as well.0960

And that is the perpendicular transversal theorem.0967

Now, if the two lines are not parallel (let's say like this), and then I tell you that this line is perpendicular to this line,0969

it is not going to be perpendicular to this line, because these lines are not parallel.0983

In this case, don't assume that it is perpendicular to both--that is only if the lines are parallel.0990

Let's do a few examples: State the postulate or theorem that allows you conclude that angle 1 is congruent to angle 2.0999

Now, remember: the only postulate was the corresponding angles one: that is the one where you have the angles in the same corner,1007

in the same position, in the same corner of the intersection--that is the corresponding angles postulate.1015

Everything else--the consecutive interior angles theorem, the alternate interior angles theorem,1026

the alternate exterior angles theorem--those are all theorems; so the only one is the corresponding angles postulate.1034

Here, what postulate or theorem allows you to conclude that angle 1 is congruent to angle 2?1043

We know that this is our transversal line, because it is the one that cuts through two or more lines.1051

Then, angle 1 and angle 2 are alternate exterior angles.1056

Now, if these two lines are parallel, then we can conclude that angle 1 is congruent to angle 2; let me show that these two lines are parallel, too.1063

Then, this would be the alternate exterior angles theorem.1073

And this one right here--we know that these are corresponding angles.1097

And the only way that the postulate will make them congruent (the only way we can apply the postulate) is if these two lines are parallel, which they are.1106

So, I can say that, by the corresponding angles postulate, angle 1 is congruent to angle 2.1114

All right, the next one: In the figure, line e is parallel to line f.1134

So, let me show this; it doesn't matter which way--I can just do like this, or I can just do like that.1142

AB is parallel to CD, so this one is parallel to this; and the measure of angle 1 is 73.1149

I am going to write that in blue; so this is 73, right here.1158

Find the measure of the numbered angles.1165

All of the numbered angles is what it is asking for.1169

Let's look at this: to look for the measure of angle 2, I know that angle 1 and angle 2 are supplementary, because they are a linear pair.1175

They form a line, and a line is 180 degrees.1190

So, all linear pairs are supplementary; so since linear pairs are supplementary, and these are a linear pair,1196

I can say that 73 plus the measure of angle 2 equals 180.1208

And then, to find the measure of angle 2, I have to subtract the 73; so the measure of angle 2 equals 107.1218

And then, the next one: the measure of angle 3--well, if you look at this, we know that these two lines are parallel.1235

This line intersects both of the parallel lines; so this is a transversal--this line segment AB is a transversal,1246

which means that angle 2 and angle 3 are alternate interior angles.1256

And by the alternate interior angles theorem, since the lines are parallel, we know that these angles are congruent.1263

Since the measure of angle 2 is 107, I can say that the measure of angle 3 is 107.1272

And then, the measure of angle 4: it is also alternate interior angles with angle 1, so by that theorem, again,1284

since the two lines are parallel, those two will be the same; so it is 73.1303

Then, the measure of angle 5 is corresponding with angle 5; angle 5 and angle 1 are corresponding,1314

because it is as if I extend this line segment, just to help me out here: these two lines are parallel;1324

here is my transversal; can you see that?--this is a line, and this is a line; here is that transversal;1332

angle 1 and angle 5 are corresponding, so if this is 73, then the measure of angle 5 has to be 73.1344

And then, the measure of angle 6--you can say that angle 6 is also corresponding with angle 3.1354

So, if you extend this out again, there is my intersection, angle 3, and then my intersection, angle 6.1368

The measure of angle 3 is 107, so the measure of angle 6 is also 107.1378

Angle 7 is alternate interior angles with angle 6, so that has to be the same, since the lines are parallel.1386

And the two lines involved would be this line and this line--can you see that?--this line and this line, and here is my transversal.1399

These two lines are parallel, so angle 6 and angle 7 are congruent by the alternate interior angles theorem.1408

And then, the last one, the measure of angle 8: it is supplementary with angle 6, because it is a linear pair.1421

Or it is alternate interior angles with angle 5, or it is corresponding with angle 4; there are a lot of different relationships going on here.1431

If you want to use the alternate interior angles theorem with angle 5 and angle 8, then it is going to be 73.1447

If you want to look at the corresponding angles postulate with angle 4, then it is also 73.1457

If you want to say that it is supplementary with angle 6 (it is a supplement to angle 6), then it is 180 - 107, which is 73.1463

You can look at it in many different ways.1475

That is it: see how all of the angle measures are either 73 or 107.1482

Since all of these lines are parallel--these pairs are parallel, and those two pairs of lines are parallel--1488

they are going to have only two different numbers, because all of their relationships are congruent or supplementary.1498

So, it is either going to be congruent, or it is just going to be a supplement to it.1507

Another example: BC is parallel to DE (that is already shown); the measure of angle 1 is 61 (this is 61);1514

the measure of angle 2 is 43; and the measure of angle 3 is 35.1527

This one is going to be a little bit more difficult, because we have lines that are closing in on the sides.1539

And sometimes it is going to be a little confusing, or a little bit hard to see the lines that you need to see.1549

And you are going to have to ignore these.1561

So, look at angle...let's see...3 and angle 4; if you look at BE as a transversal, and these two1564

as the lines that the transversal is intersecting, 3 and 4 are alternate interior angles.1584

But these two lines are not parallel--those two lines that the transversal is intersecting are not parallel.1594

So, you can't assume that they are congruent; you can't say that they are congruent, because look: the two lines are intersecting.1602

Even though they are alternate interior angles, you can't apply the theorem saying that they are congruent, because the lines are not parallel.1612

You have to be very careful; you can't say that angle 4 is 35 degrees.1620

OK, so what can we say? We know that this line segment right here is parallel to this line right here.1626

I can say that the measure of angle 5...because look at this: this angle 5 and angle 2 are alternate interior angles;1641

now, let's see if we can apply the theorem and say that they are congruent.1658

Here is my transversal; here are the two lines that the transversal is intersecting; are the two lines parallel?1663

Yes, they are parallel; now, ignore this side and this side, AD and AE, because you don't need them.1671

It is as if they are not even there; cover it up.1681

Angle 5 doesn't involve those lines; angle 2 doesn't involve those lines.1684

So, all you have to see is this right here; here is BC; there is angle 5 and angle 2.1690

Here, these are parallel; here is 5, and here is 2.1700

So then, these are alternate interior angles, and they are congruent, because their lines are parallel.1708

The measure of angle 5 would be 43.1716

And then, from here, I can say that the measure of angle 7...if you look at angle 7 and angle 1, I have a transversal;1721

there is my angle 7, and there is my angle 1; these two lines are parallel.1752

See how it is only involving the three lines, this line, this line, and this line.1760

Ignore BE; see how I didn't draw it, because it is not involved.1766

Ignore all of the other lines; just look at those three lines for angle 1 and angle 7.1769

If it helps, you can draw it again; this one is a little bit hard to see using this diagram,1774

so if it helps you like this, then just draw it again, just using those three lines.1779

Angle 7 and angle 1 are corresponding; and since the lines are parallel, I can use the postulate to say that angle 1 and angle 7 are congruent.1785

So, the measure of angle 7 is 61.1794

So then, the ones that I found: this is 43; this is 61.1799

OK, to find the measure of angle 4, I can say that, because all these three angles right here form a linear pair,1807

that the measure of angle 7 plus the measure of angle 5 plus the measure of angle 4--they are all going to add up to 180,1824

because they form a straight line; all three angles right here are going to form a 180-degree angle.1833

You can say that the measure of angle...not 1....4, plus 61, plus 43, equals 180.1844

The measure of angle 4, plus 104, equals 180; you subtract the 104, so the measure of angle 4 equals 76; here is 76.1861

All of this is the one that I found, so I will write this in red: 76.1890

And then, let's look at some other ones: now, if you look at angle 8, angle 8 also involves this parallel line.1895

But this one is a little bit harder to see, because you have angle 8 like that; what is this angle right here?1915

This is angles 2 and 3 together; it is this angle and this whole thing.1930

Now, ignore this line; you are just involving this line, this transversal, and this bottom line DE.1936

So, this BE is not there; so it would just be this whole angle together.1946

So then, see how this angle right here and this angle right here are corresponding.1954

But this has another line coming out of it like this to separate it into angles 2 and 3.1961

All I have to do is add up angles 2 and 3, and that is going to be my angle 8.1970

This is going to be 78 degrees, and since the lines are parallel, the corresponding postulate says that they are congruent; that equals 78.1975

I will write that here: 78; and then, this 78 and angle 6 are going to form a linear pair.1993

Right here, 78 +...now, since this is angle 6 right here, you can look at angle 6 and this angle right here,2013

78 degrees, because that is angles 2 and 3 combined; they are going to be consecutive interior angles.2029

And they are supplementary; so you can just do angle 6 + 78 = 180, which is the same thing as looking at this.2038

These are supplementary, so angle 6 and angle 8 (78) are going to add up to 180.2049

It is the same thing: the measure of angle 6, plus 78, is going to equal 180.2055

The measure of angle 6: if I subtract 78, then you get 102, so this is 102.2070

Now, with angle 9, to find the measure of angle 9, that is actually going to involve using the triangles,2088

because the only relationship that this angle has with any of the other angles is that it forms within the triangle.2099

And see how angle 9 is not supplementary; it doesn't form a linear pair; there is no transversal involved with angle 9.2111

It is just these two angles, or those two right there.2120

Angle 9 is actually going to involve what is called the triangle sum theorem, where the three angles of a triangle are going to add up to 180.2125

So, we haven't gone over that yet; if you want, you can just say that the measure of angle 9, plus 61 (this angle),2134

plus the 78, is going to equal 180, and then find the measure of angle 9 that way.2146

You can also look at this big triangle and say that this angle, plus this angle, plus this angle, are going to add up to 180.2151

You can also look at it as this triangle right here, saying the measure of angle 9 plus 75, together, and then 3, are going to be 180.2163

And then, find the measure of angle 9 that way.2175

So, for now, we are just going to solve for these; and that is it for this problem.2179

The last example: Find the values of x, y, and z.2186

Here you have three lines: now, these three lines are going to be parallel.2192

I am going to make them parallel, so that I can solve for these values.2199

Now, the only angle that is given is right here, 118.2203

If you look at this, again, we have four lines involved; and to form special angle relationships, you only need three lines.2209

You need the transversal and the two lines that it intersects to form those pairs of angles.2219

Whichever lines you are using, always keep them in mind, and then look at what line you are not going to use,2228

and ignore that line, since we have four and we only need three.2238

Using this angle right here, 118, I can say that now this one right here and 11z + 8 are corresponding.2245

And then, this one right here and this one right here are alternate exterior angles, because it is involving these three lines, and not this one right here.2262

These would be alternate exterior angles.2273

Or, if I ignore this middle line, and I just say that this transversal with this line and this line2276

(again, ignoring the middle line--pretending it is not there), then 118, this angle right here, with x, would be alternate exterior angles.2290

Imagine if you have a line, a line...here is your transversal; the middle line is not there; this is x, and then this is 118.2303

You see that it is alternate exterior angles.2314

So then, I can say that x is equal to 118, because the lines are parallel.2319

And so then, I can apply the alternate exterior angles theorem, saying that that relationship, that pair, is congruent.2325

The next one: let's look at z; this one right here, 11z + 8, is going to equal 118.2337

Why?--because, if I look at this line, with this line and this transversal, they are going to be corresponding angles.2346

And then, since the lines are parallel, the corresponding angles postulate says that they are congruent.2358

11z + 8 = 118; so if you subtract the 8, 11z = 110; z = 10.2366

There is my x; there is my z; and then, I have to find y now.2387

For my y, I can say that this angle with 118--they are not congruent, remember, because they are going to form a linear pair.2392

They form a line, so they are going to be supplementary.2407

You can also note that this angle is 118, remember, because we said that they were corresponding--this one with this one.2412

So, since this is 118, this angle with this angle would be consecutive interior angles.2421

And if the lines are parallel, then the theorem says that they are supplementary, not congruent.2430

So, either way, 3y + 2 =...not 180; you have to say that this whole thing, plus the 118, is going to equal 180.2436

3y + 2 = 62, and then, if you subtract the 2, then 3y is going to equal 60; y is going to equal 20.2456

x is 118; y is 20; and z is 10; just remember to keep looking for those relationships between the pairs.2474

You can also definitely use the linear pair, if they are supplementary; you can definitely use that.2482

If they are vertical, definitely use that, because you know that vertical angles are congruent.2490

So, any of those things--you have a lot of different concepts that you learn that will help you solve these types of problems.2499

That is it for this lesson; thank you for watching Educator.com.2510

Hello--welcome back to Educator.com.0000

The next lesson is on the slope of lines; this might be a little bit of a review for you from algebra.0002

But this whole lesson is going to be on slope.0010

Slope is the ratio between the vertical and the horizontal, or we can say "rise over run."0014

Rise, we know, is going up and down; and then, the run is going left and right.0027

So, when we talk about slope, we are talking about the vertical change and the horizontal change.0033

For slope, we use m; so the slope of a line containing two points with coordinates (x1,y1)0044

and (x2,y2) is given by this formula right here.0057

Now, (x1,y1), these numbers right here, and (x2,y2), those numbers, are very different from exponents.0061

They are not to the power of anything; it is just saying that it is the first x and the first y.0073

So, we know that all points are (x,y), and so, this right here is just saying that this is the first x and the first y.0083

And this is also x and y, but you are just saying that it is the second point; it is the second x and the second y.0094

Because you have two x's and two y's, you are just differentiating the points; this is the first (x,y) point, and this is the second (x,y) point.0103

It doesn't matter which one you label as first and which one you label as second.0112

You are just talking about two different points.0116

And when you have two points, then the slope is going to be (y2 - y1)/(x2 - x1).0118

You are going to subtract the y's, and that is going to be your vertical change, because y, you know, is going up and down.0130

And x2 - x1 is your horizontal change, the difference of the x's, which is going horizontally.0139

Now, it doesn't matter...like I said, if you are going to subtract this point, this second y, from this y,0150

then you have to make sure that you subtract your x's in the same order.0163

If you are going to subtract y2 - y1, then it has to be x2 - x1 for the denominator.0168

It can't be (y2 - y1)/(x1 - x2).0174

If you do y2 - y1 over here, you cannot do x1 - x2.0179

You can't switch; it has to be subtracted in the same order, or else you are going to get the wrong answer.0185

And this right here is just saying that x1 and x2, these numbers, can't equal each other,0190

because if they do, then this denominator is going to become 0.0197

If x1 is 5, and x2 is 5, then it is just going to be 5 - 5, and that is going to be 0.0201

And when we have a fraction, you can't have 0 in the denominator, or else it is going to be undefined.0212

So, that is what it is saying right here: they should not equal each other, or else you are going to have an undefined slope.0218

Let's find the slope of these lines: here we have (-4,-2) and (5,3).0229

Now, you can do this two ways: you can use the slope formula by doing (y2 - y1)/(x2 - x1);0239

if you have a coordinate plane, and it is marked out for you--you have grids that show each unit--0252

then you can count; you can just go from one point to the other point,0262

and you can just count your vertical change and count your horizontal change; you could do it that way.0269

But since these are not labeled--each unit is not labeled out--let's just use the slope formula.0275

Here, if I make this (x1,y1), (x2,y2), then my slope is going to be (-2 - 3)...0285

so then, this value is (y2 - y1, which is 3), over (-4 - 5).0301

Now, I could do (3 - -2); I can go that way if I want, but if I do that, if I choose to do this one first,0312

(3 - -2), then I have to do (5 - -4); you have to be in the same order.0321

If you do 3 minus -2, then you can't go with (-4 - 5); you can't go the other way then.0327

It doesn't matter which one you start with; but when you do your x, you have to do it in the same order.0338

This one right here is -5/-9, which is just 5/9.0345

Now, without solving slope, if you look at the line, you should be able to tell if the slope is going to be positive, negative, 0, or undefined.0358

For this one, since the slope measures how slanted a line is, how tilted a line is, if we look at this line,0374

imagine a stick man (I like to call him "stick man," because I can only draw stick figures) walking on this line.0388

Now, he can only walk from left to right, because let's say you read this--you have to read from left to right.0398

So then, it can only go from left to right; he is walking uphill, and that would be a positive slope.0407

This is a positive slope; if the stick man is walking uphill, it is a positive slope.0416

If the stick man is walking downhill, like the next one (again, he can only walk from left to right)--he is going to walk downhill, so this is a negative slope.0426

Without even solving, I know that my slope is going to be negative.0439

This is positive; it is positive 5/9.0443

Now, the slope for this one--I know, before I even solve it, that it is going to be negative.0446

So, after I do solve it, if I get a positive answer, then I know that I did something wrong, because it has to be negative.0451

For this one, the slope is 5 - -4; and make sure that you are going to find the difference of the y's for your numerator.0463

Don't do your x's first; the numerator is y's; the denominator is x's.0475

I went from this to this, so then I have to do -2 - 3.0482

So then, this is...minus a negative is the same thing as a plus, so 5 + 4 is 9, and then -2 - 3 is -5.0489

So, this is -9/5; and I have a negative slope, so that is my answer.0502

A couple more: here I have a horizontal line; my slope is (y2 - y1)...(-3 - -3), over (-6 - 4); this is 0,0519

because -3 + 3 is 0; this is -10; well, 0 over anything is always 0; so the slope here is 0.0541

Now, let's bring back the stick man: if stick man is walking on this, he is not walking uphill or downhill; he is just walking on a flat surface.0554

If he is walking on a flat surface, since slope measures how slanted a line is, it is not slanted at all--it is just flat; that is why the slope is 0.0563

Whenever it is flat, it is a horizontal line, and the slope will be 0--always.0573

It doesn't matter if it is up here or down here; as long as it is a horizontal line, your slope is going to be 0.0580

The next one: 4 - -4...be careful with the negatives: it is 4 minus -4; -2 - -2; change those to a plus--0592

minus a negative is also a plus--so 4 + 4 is 8, over -2 + 2...is 0.0615

Now, look at the difference between this one and this one.0624

In this one, the 0 is in the numerator; if it is in the numerator, it makes it just 0; 0 is a number, just like 5, 6, -3;0628

those are all numbers, and 0 is a number; so your answer for this problem, your slope, is 0; you have a slope; it is 0.0640

And in this problem, we cannot have 0 in the denominator--it is just not possible.0650

So, since you did come up with a 0 in the denominator, this is going to be undefined.0657

You can also say "no slope"; in this case, you do have a slope--the slope is 0; in this case, you do not have a slope.0669

There is no slope; it doesn't exist; it is undefined, because the denominator is 0.0680

So, my answer is just "undefined."0685

And then, to bring back the stick man: since it is a man, it can't do this--this is like walking up a wall.0689

Stick man can't walk up a wall; it is not possible--it can't do it.0702

He can't walk up a wall; he is not Spiderman; he can't walk up a wall.0706

So, in this case, this man can't do this; if he can't walk up this wall, it is undefined; it can't be done; there is no slope.0711

If he is walking on a horizontal--no slant at all--it is 0.0724

If he has to walk up a wall (which is impossible), then it is an undefined slope.0731

If the stick man is walking uphill, it is a positive slope; downhill is a negative slope; a horizontal line is 0; a vertical line, like a wall, is undefined.0737

You can't walk up a wall.0749

A couple of postulates: If we have two non-vertical lines that have the same slope, then those lines are parallel,0754

because again, slope measures how slanted a line is.0769

So, if I have two lines that are slanted exactly the same way, then they are going to be parallel.0775

Again, two lines that have the same slope are parallel.0788

And this part right here: "if and only if"--now, we went over conditionals, if/then statements;0794

to change this one right here (let's go over this...number 1)...two non-vertical lines have the same slope if and only if they are parallel.0807

This just means that this conditional and its converse are both true.0819

It is basically two statements, two conditionals in one.0833

I can say, "If two lines have the same slope, then they are parallel."0840

And this would be the converse: I can say, "If two lines are parallel, then they have the same slope."0865

The statement and its converse are both true: this is true, and this is true.0893

So, just instead of writing each of those conditionals separately, you can write them together by "if and only if."0897

It just means that this statement and its converse (converse means, remember, that you switch the hypothesis and the conclusion) are both true.0908

So then, you can just use "if and only if."0919

If two lines have the same slope, then the two lines are parallel.0925

Or you can say, "If two lines are parallel, then they have the same slope."0930

Either way, parallel means same slope; same slope means parallel.0934

Now, the next postulate is talking about perpendicular lines: Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.0939

Now, "the product of their slopes is -1"--that means that, if, first of all, I have a line like this,0957

and I have a line like this, let's say they are perpendicular; and the slope of this line, let's say, is 1/20970

(it has to be positive; it is going uphill); then the slope of this line is going to be the negative reciprocal,0979

meaning that you are going to make it negative; if it is negative already, then you are going to make it positive;0990

so, the slope of this line will be negative...and then the reciprocal of it will be 2/1.0997

So then, it is saying that the product of their slopes is going to be -1; so 1/2 times -2 is -1.1006

Just think of it as: If you have two perpendicular lines, then the slopes are going to be negative reciprocals of each other.1023

And if you multiply those two slopes, then you should get -1, always.1032

OK, parallel or perpendicular lines: you are given points A, B, C, and D; you want to determine if line AB is parallel or perpendicular to CD.1040

So, to determine if the two lines are parallel or perpendicular, then you have to compare their slopes.1058

So, for line AB, I need to use points A and B.1066

If I find the slope using these two points, the slope of AB is going to be -6 - 0, y2 - y1, over x2 - x1.1070

And that is -6/-3, so we have 2.1093

And then, the slope of CD is y2, -3, minus -4, over 4 - 2; this is 1/2, so this is positive 1 over 2.1100

Did I get that right?--yes.1133

In this case, it is going to be neither, because here we have AB; (-6 - 0)/(-2 - 1) becomes positive 2.1138

And then here, this is -3 - -4, and 4 - 2; for this, I get positive 1/2.1157

Now, it looks like they are going to be perpendicular, but remember: they have to be the negative reciprocal of each other.1168

If this is 2, this is 2/1, and the inverse, or the reciprocal, is 1/2; but they are not negative--it is not negated.1176

So, if I multiply 2/1 times 1/2, I am only going to get 1, not -1; so this is neither.1188

OK, find the value of x if the line that passes through this point and this point is perpendicular to the line that passes through (-1,6) and (-2,8).1204

We are given our two points that we have to find the slope of.1220

But from those two points, one of the values, the x-value, is missing.1227

That means that we need the slope.1235

They didn't just give us the slope in this problem; they didn't just hand it to us.1239

We have to actually solve for the slope, because we know that it is perpendicular to a line that passes through these two points.1243

So, basically, the points that I have to use are (x,4) and (-3,3).1255

I have to find x; and this is another line, and I am just going to use that line to find the slope,1266

because I have my slope that I need that has a relationship with the slope of this line.1278

So, to find the slope of this line right here that passes through these points,1285

I am going to do (6 - 8), (y2 - y1), over (-1 - -2); this is -2/1, so the slope is -2.1293

But since I know that my line is perpendicular to this line, my slope is going to be the negative reciprocal of this slope.1307

If this slope is -2, then my slope is going to be positive 1/2, positive one-half.1322

That is what I need to use: the slope is positive 1/2.1333

Now, using the slope formula, I know that this is (y2 - y1)/(x2 - x1).1338

Well, I can just fill everything in, except for this, and then use that as x1.1350

1/2 is my m; y2...if this is (x1,y1), (x2,y2),1356

y2 is 3, minus 4, over -3, minus x; since I don't know this value, which is what I am solving for, I can just leave it like that.1365

And then, from here, I have to solve this out.1384

I can solve this out a couple of ways: first of all, since this is a fraction equaling a fraction, I can use proportions.1388

I can make (-3 - x) times 1 equal to 2 times 3 minus 4; or let me just do this--let me just simplify this first.1399

1/2 = -1/(-3 - x); or I can just multiply...1411

I have a variable; the variable that I am solving for is x, and that is in the denominator.1423

If I want to solve for the variable, it cannot be in the denominator, so I have to move it out of the denominator.1427

I can do that by multiplying both sides or the whole thing by -3 - x;1433

or again, since this is like a fraction equaling a fraction, like a proportion, I can just do that.1439

So, just make -3 - x equal to...and then, I am just multiplying it this way...equaling this; it is -3 - x = -2;1444

if I add 3, then I get -x = 1; x = -1.1459

So again, you are going to find your slope.1469

They might not just hand you the slope; they might not tell you what the slope is directly.1476

So then, you have to find it this way; they will give you another line that has a relationship with your line, your slope.1483

So, you have to find the slope of that other line, and then use that slope to find your slope.1496

And then, plug it all into the slope formula; and then from there, you just solve.1503

Let's do a few examples: Find the slope of the line passing through the points.1510

Again, here is the slope formula; this equals...it doesn't matter which one I use first, so I will just use (-2 - 5) first.1517

That means that I have to use this one first; so it is (3 - -4).1536

This becomes -7/7, which is equal to -1; and then, for this one, the slope is (0 - 6)/(-7 - -7).1542

This is going to be -6/0; 0 is in the denominator, which means that I have an undefined slope.1564

And that just means that the line that is passing through these points is going to be a vertical line; vertical lines have undefined slopes.1580

Find the slope of each line: now, they don't give you the points--they just show me the lines.1593

And I have to see what points the lines are going through to find the slope.1602

Let's see, let's look for the slope of n first; here is n.1612

Now, remember: for slope, I can do this two ways: I can find two points on this line, like this and like this--1618

those are two points on the line (or here is another point; it doesn't matter--any two points on the line);1630

you can find the coordinates of the points and use the slope formula.1636

For two points, find the coordinates and use the slope formula.1642

Or an easier way, in this problem: Since we have all of these grids marked out for us, I can just1646

(because slope is rise over run, the vertical change over the horizontal change; rise is how many it is going up or down,1656

and then run is how many is going left or right)--whenever I go up (here, this is the positive y-axis), any time I am counting upwards,1668

it is a positive number; if I am counting downwards, then it is a negative, because I am going smaller.1683

If you are going up, it is a positive number; if you are going down, if you are counting down, then it is a negative number.1690

The same thing for x: if you are moving to the right, it is a positive number; if you are moving to the left, it is a negative number,1696

because it is getting larger as you go to the right; and as you go to the left, you are moving towards the negative numbers.1703

You are getting smaller, so it is negative if you are going to the left.1709

To find the slope of n, I am just going to do rise over run; I am going to just count my vertical and horizontal change.1715

You go from any point to any other point on the line.1724

I can start from here; I am going to go one up, because it is on this right here.1729

My vertical change: I only went up one; remember, up is positive, so to find the slope of n, it is positive 11740

(that is my rise), and then I am going 1, 2 to the right--that is positive 2, so the slope is 1/2.1754

Now, remember: I can also go from any point to any other point on the line.1767

So, if I start from this point, let's say I am going to go from this point, and then (I didn't see this point, so) to this point;1773

then I can go down 2 (remember: down 2 is negative 2), over...then I am going to go 1, 2, 3, 4.1781

And that is to the left, so that is negative 4; -2/-4 is the same thing as 1/2.1797

It doesn't matter how you go from whichever point to any other point on the line, as long as both points are on the line,1808

and as long as you make it a positive number going up, a negative number going down, positive to go right, and negative to go left.1817

You are going to get the same answer; you are going to get the same slope.1826

The slope of n is 1/2; then the slope of p (let me use red for this one): let's see, I have a point here, and I have another point here.1830

So again, I can go from this point to that point, or I can go from that point to that point; it doesn't matter.1853

Let's start right here: I am going to go 1, 2, 3; I went up 3, so that is a positive 3.1858

And then, from here, I go to the left 1, which is a negative 1.1869

3/-1 is the same thing as -3, so my slope of p is -3.1876

Or I could go from this point to this point; that would be to the right one (that is positive 1), over down 3 (1, 2, 3);1883

oh, I'm sorry; I did horizontal over vertical, which is wrong; so I have to go this way--vertical first.1893

1, 2, 3: that is a -3, and to the right 1--that is positive 1; so this is also -3, the same thing.1902

The next line is line q (I will use red for this one, too): this is a vertical line.1916

Automatically, I know that that has an undefined slope. I can also just...1932

Now, I know that this line is not really completely lined up with the grid, but sometimes when you transfer1939

this into this program, or move things into this program, it might shift a little bit.1947

But think of this line as being on 2 right here, as x being 2.1951

Let's say I have this point right here, and then any other point--that point right there.1960

All that I am doing is: my vertical change is going down to -2; my horizontal is nothing: 0.1964

I am not moving to the right or left at all; that is 0.1975

So, we have a 0 in the denominator; this is an undefined slope.1979

And again, I knew that because this is a vertical line; the stick man can't walk up that line; so it is an undefined slope.1989

And the last one, for line l: any two points...1999

Again, this line is shifted a little bit, but I can just do that if I want.2007

Vertical change first: the vertical change is 0, because I am not moving vertically; to get from this point to this point, I don't go up or down at all.2014

So, it is 0 over...and then, I can move 1, 2, 3 to the right; so no matter what the bottom number is, my slope will be 0.2021

The slope of the l is 0; again, it is a horizontal line, so it is not slanted; it is not going uphill or downhill--nothing.2035

It is just horizontal; then the slope is 0.2044

The next example: Graph the line that satisfies each description; slope is 2/3 and passes through (-1,0).2052

You just have to graph this first one; let's say I am going to graph it right up here.2066

Now, just a sketch will do; let's say...here is my x; here is my y; (-1,0) is right there.2075

My slope is 2/3; so again, this is rise over run.2103

Now, I can use the same concept, the positive going up and negative going down, positive to the right and negative to the left.2110

The top number, the rise, to go up and down: I have a positive 2--that means I am going to go up 2, because it is positive.2120

From here, I am going to go 1, 2; and then, I have 3 that I am going to move to the left or to the right;2130

but since it is a positive 3, I am going to move to the right: 1, 2, 3.2139

Now, from this point, I can go down if I want to, because 2/3, that slope, is the same thing as -2/-3.2150

So, if I go -2, I am going to go down 2: 1, 2; and then, -3 is to the left, so 1, 2, 3; there is my line, right there.2163

This is the second one: it passes through point (3,1) and is parallel to AB with A at this point and B at that point.2182

Again, they don't give us our slope; they just give us the point that we need to use.2194

Our line is going through this point, and we don't have the slope of our line;2201

instead, they give us the slope of another line, line AB; and they say that it is parallel to it.2205

So, as long as we find the slope of AB, since it is parallel, we know that our slope will just be the same as this slope.2211

The slope of AB is (4 - 3)/(-1 - -2), which is 1 over...this is 1...so 1.2222

Now, since our slope is parallel, again, our slope is 1.2238

And then, this is our point; so we have point (3,1), and the slope is 1.2247

To graph (x,y), (3,1), it is 1, 2, 3; and 1; there is our point that our line is passing through.2258

And then, our slope is 1; 1 is the same thing as 1/1, so positive 1 is up 1, to the right 1; also, positive 1 is up 1.2291

And then again, you can just do -1/-1; that is the same thing as 1.2308

So, from here, I can go down 1, left 1; down 1 is -1; left 1 is -1.2313

And that is going to be a line like that.2322

The last example: Determine the value of x so that a line through the points has the given slope.2331

Again, they give us a slope, and then we have to find the missing value, which is x.2338

Since we know that the slope formula is (y2 - y1)/(x2 - x1),2345

if I make this (this is (x,y), and this is also (x,y)) my first point, and this is my second point--2354

so that is (x2,y2)--then my slope is y2, which is -5, minus 1, over x2, -3, minus x.2366

Now, again, you can turn this into a proportion; or I can just multiply this out to both sides.2385

I can multiply this to this right here to cancel it out, and then multiply to the other side and distribute that; I could solve it that way.2397

Or I could just make this a proportion: 2/1 = -6/(-3 - x); so to continue right here, it is going to be 2(-3 - x) = -6.2406

This is -6 - 2x, and you just distribute that; that equals -6; -2x = 0; x = 0.2430

So, we have 0 as our x for this problem.2444

OK, the next one: again, plugging everything into the slope formula, we have 4/3 = (-2 - -6)/(x - 7).2450

So, 4/3 =...this is -2 + 6, so this is 4, over x - 7.2472

OK, well, in this problem, again, you can multiply x - 7 to both sides to get it out of the denominator,2482

because since you are solving for x, it cannot be in the denominator.2492

Or you can cross-multiply using proportions, because it is a fraction equaling a fraction.2497

Or, since this 4 numerator equals this 4 numerator, then the denominator has to equal the denominator, so you can make 3 equal to x - 7.2505

So, let's just solve it out, multiplying: I can multiply this side by (x - 7) like that,2518

and multiply this side by (x - 7); then, it is going to be 4/3x minus 28/3 equaling 4.2529

Now, this is actually probably the harder way to do it; but I just wanted to show you how to multiply it to both sides.2553

This is a binomial that, again, you are just multiplying to both sides.2561

But the easiest way would be to make (x - 7) equal to 3: because 4 = 4 (the top), then 3 = x - 7 there.2567

So, let's just continue out this way: if I add 28/3 to both sides, this is going to be the same thing as 12/3 + 28/3,2577

and that is just because I need a common denominator; that is equal to 4/3x.2593

That equals...this is 40/3...if I multiply the 3's to both sides, this will be 4x = 40; x = 10.2605

So here, x = 10, this value right here.2626

And if we solved it the other way, 3 = x - 7, then you would just add 7, so x would be 10.2632

That is it for this lesson; thank you for watching Educator.com.2642

Welcome back to Educator.com.0000

This next lesson is on proving lines parallel.0002

We are actually going to take the theorems that we learned from the past few lessons, and we are going to use them to prove that two lines are parallel.0007

We learned, from the Corresponding Angles Postulate, that if the lines are parallel, then the corresponding angles are congruent.0022

If the lines are the parallel lines that are cut by a transversal, then the corresponding angles are congruent.0035

Now, this one is saying, "If the two lines in a plane are cut by a transversal, and corresponding angles are congruent, then the lines are parallel."0045

So, it is using the converse of that postulate that we learned a couple of lessons ago.0053

And we are going to take that and use it to prove that lines are parallel.0058

Before, when we used that postulate, it was given that the lines were parallel; and then you would have to show0065

that the conclusion, "then the corresponding angles are congruent," would be true.0071

But for this one, they are giving you that the corresponding angles are congruent.0077

And then, the conclusion, "the lines are parallel," is what you are going to be proving.0083

This first postulate: if you look at angles 1 and 2, those are corresponding angles.0090

So, if I tell you that angle 1 and angle 2 are congruent, and they are corresponding angles, then the lines are parallel.0096

As long as these two angles are congruent, then these lines are parallel; so I can conclude that these lines are parallel.0104

Now, again, for them to be corresponding angles, the lines don't have to be parallel.0115

Even if the lines are not parallel, they are still considered corresponding angles.0124

But now, we know that, as long as the corresponding angles are congruent, then the lines will be parallel.0129

The next postulate: If two lines in a plane are cut by a transversal so that a pair of alternate exterior angles are congruent, then the two lines are parallel.0136

So again, this is the alternate exterior angles theorem's converse.0149

The alternate exterior angles theorem said that, if two lines are parallel, then alternate exterior angles are congruent.0156

This is the converse; so they are giving you that alternate exterior angles are congruent; then, the lines are parallel.0168

Depending on what you are trying to prove, you are going to be using the different postulates--either the original theorem or postulate, or the converse, these.0176

If you are trying to prove that the lines are parallel, then you are going to be using these.0188

If you are trying to prove that the angles are congruent, then you are going to be using the original.0191

So, alternate exterior angles are congruent; therefore, we can conclude that the lines are parallel.0197

The Parallel Postulate: this is called the Parallel Postulate: If there is a line and a point not on the line,0209

then there exists exactly one line through that point that is parallel to the given line.0220

What that is saying is that I could only draw one line, a single line, through this point, so that it is going to be parallel to this line.0227

I cannot draw two different lines and have them both be parallel to this line--only one.0239

So, it would look something like that...well, that is not really through the point, but...0246

this is the only line that I can draw to make it parallel to this line.0256

Only one line exists; and that is the Parallel Postulate.0262

Now, we are going to go over a few more theorems now.0270

Before it was postulates, but now these are some theorems that we can use to prove lines parallel.0273

If two lines in a plane are cut by a transversal (this is my transversal) so that a pair of consecutive interior angles is supplementary, then the lines are parallel.0281

Remember consecutive interior angles? They are not congruent; they are supplementary.0292

If you look at these angles, angle 1 is an obtuse angle; angle 2 is an acute angle.0297

They don't even look congruent; they don't look the same.0303

Make sure that consecutive interior angles are supplementary.0307

Again, the original theorem had said, "Well, if the lines are parallel" (that is given), "then we can conclude that consecutive interior angles are supplementary."0312

This one is the converse, saying that the given is that the consecutive interior angles are supplementary.0323

Then, the conclusion is...we can conclude that these lines are parallel.0330

If two lines in a plane are cut by the transversal, so that a pair of alternate interior angles is congruent, then the lines are parallel.0338

So again, if these alternate interior angles are congruent, then the lines are parallel.0350

And make sure that it is not the transversal; it is the two lines that the transversal is cutting through that are parallel.0363

And the next theorem, the last theorem that we are going to go over today for this lesson:0371

in a plane, if two lines are perpendicular to the same line, then they are parallel.0375

See how this line is perpendicular to this line? Well, this is my transversal.0383

So, if this line is perpendicular to this line, and this line is also perpendicular to that same line, the same transversal, then these lines will be parallel.0387

If both lines are perpendicular to the same line, the transversal, then the two lines will be parallel.0405

OK, let's go over a few examples: Determine which lines are parallel for each.0418

This one right here, the first one, is giving us that angle ABC is congruent to angle (where is D?...) DGF.0426

That means that this angle and this angle are congruent.0440

OK, now again, when we look at angle relationships formed by the transversal, we only need three lines.0445

We have a bunch of lines here; so I want to just try to figure out what three lines I am going to be using for this problem,0456

and ignore the other lines, because they are just there to confuse you.0466

This angle right here is formed from this line and this line, so I know that those two lines, I need.0474

And then, this angle is also formed from this line and this line.0480

So, it will be line CJ, line FN, and line AO; those are my three lines.0484

This line right here--ignore it; this line right here--ignore it.0492

We are only dealing with this line, this line, and this line.0495

And from those three lines, we know that this line, AO, is the transversal, because that is the one that is intersecting the other two lines.0499

So, if this angle and this angle are congruent, what are those angles--what is the angle relationship?0509

They are corresponding; and the postulate that we just went over said that, if corresponding angles are congruent, then the lines are parallel.0517

I can say that line CJ is parallel to line FN.0530

The next one: angle FGO, this angle right here, is congruent to angle NLK.0546

So again, I am using this line, this line, and this line, because it is this angle right here and this angle right here.0557

So, for those two angles, their relationship is alternate exterior angles.0567

If alternate exterior angles are congruent, then the two lines are parallel.0575

And those two lines are going to be, since this FN is a transversal, line AO, parallel to HM.0583

And remember that this is the symbol for "parallel."0598

The measure of angle DBI (where is DBI?), this angle right here, plus the measure of angle BIK0604

(BIK is right here) equals 180, so that these two angles are supplementary.0614

Now, these two angles are consecutive interior angles, or same-side interior angles,0621

which means that if they are supplementary (which they are, because 180 is supplementary), then the two lines are parallel.0629

That is what the theorem says; so I know that line AO, just from this information, is parallel to HM.0637

The theorem says that, if the two angles are supplementary, then the two lines are parallel.0654

OK, the next one: CJ is perpendicular to HK; this is perpendicular, this last one.0661

And then, FN is perpendicular to KN; there is the perpendicular sign.0671

Remember the theorem that said that, if two lines are perpendicular to the same transversal, then the two lines are parallel.0677

So then, from this information, I can say that CJ is parallel to FN.0688

The next example: Find the value of x so that lines l and n will be parallel.0703

I want to make it so that my x-value will make them have some kind of relationship, so that I can use the theorem,0715

saying that I have to make two angles congruent or supplementary--something so that I can conclude that the lines are parallel.0728

Let's see: these two angles right here don't have a relationship; this one is an interior angle, and this one is an exterior angle.0742

But what I can do is use other angle relationships; if I use other angle relationships, then I can find some kind of relationship0754

from the theorems or the postulates that we just went over.0770

What I can do: since this angle and this angle right here are vertical angles, I know that vertical angles are congruent.0774

And since vertical angles are always congruent, and these are vertical angles, since this is 4x + 13,0786

I can say that this is also 4x + 13, because it is vertical, and vertical angles are congruent.0792

Now, this angle and this angle have a special relationship, and that is that they are same-side or consecutive interior angles.0801

I know that, if consecutive interior angles are (not congruent) supplementary, then the lines are parallel.0812

As long as I can prove that these two are supplementary angles, then I can say that the lines are parallel.0823

I am going to make 4x + 13, plus 6x + 7, equal to 180, because again, they are supplementary.0832

Then, I am going to solve for x; this is going to be 4x + 6x is 10x; 13 + 7 is 20; 10x = 160, so x is going to be 16.0851

So, as long as x is 16, then that is going to make these angles supplementary, and then the lines will be parallel.0870

So, x has to be 16 in order for these two lines to be parallel.0880

Find the values of x and y so that opposite sides are parallel.0889

OK, that means that I want AB to be parallel to DC, and I want AD to be parallel to BC.0894

So, the first thing is that...now, this is a little bit hard to see, if you want to think of it as parallel lines and transversals.0902

So, what you can do: if you get a problem like this, you can make these lines a little bit longer.0918

Extend them out, so that they will be easier to see.0924

That means that this one right here and this one right here--if this is the line,0939

and these are the two lines that the transversal is cutting through, then these two angles are going to be consecutive interior angles.0947

This angle and this angle are consecutive interior angles.0956

This angle and this angle are also consecutive interior angles, because it goes line, line, transversal.0959

That means that these are same-side interior angles, or consecutive interior angles.0967

Then, I have options: since I have the option of making this one and this one supplementary, I can also say that this one and this one are supplementary.0973

But this one has the y, and this one has an x; so instead of using x and y to make it supplementary0986

(you are going to have 2 variables), I want to use this one first.0996

I want to solve this way, because it has x, and this has x--the same variable.0999

And you want to stick to the same variable.1005

Here I can say 5x (and I am going to use that, if consecutive interior angles are supplementary, then the lines are parallel) + 9x + 12 = 180.1009

And then, if I do 14x + 12 = 180, 14x =...if I subtract that out, it is going to be 168.1028

Then, x equals, let's see, 12; so if x is 12, then that is my value of x, and then I have to find the value of y.1041

Now, this angle measure, then, if x is 12, will be 60, because 5 times 12 is 60.1063

Then, this right here: 9 times 12, plus 12, is going to be 120.1074

Now, I also know that this and this are supplementary, so the 60 + 120 has to be 180; that is one way to check your answer.1082

Now, remember: earlier, we said that this angle and this angle are also consecutive interior angles, which means that they are supplementary.1092

Well, if this is 60, then this has to be 120, because this angle and this angle are supplementary; they are consecutive interior angles.1099

So, since this is 120, to solve for y, I can just make this whole thing equal to 120.1109

7y + 10 = 120; subtract the 10; I get 7y = 110; and then, y = 110/7.1115

And that is simplified as much as possible, so that would be the answer.1139

Now, when you get a fraction, it is fine; it is OK if you get a fraction.1150

You can leave it as an improper fraction, like that, or you can change it to a mixed number.1153

But this is fine, whichever way you do it, as long as it is simplified1160

(meaning there are no common factors between the top number and the bottom number, the numerator and denominator).1165

Then, you are OK; there is x, and there is y.1173

You solved for the x-value and the y-value, so now that these consecutive interior angles are supplementary, I can say that these sides are parallel.1178

And then, since this one and this one, consecutive interior angles, are supplementary, this side and this side are parallel.1192

Now, notice how, for these, I did it once; and then, for these, I had to do it twice,1202

because any time you have the same number of these little marks, then you are saying that they are congruent;1210

if they are slash marks, then they are congruent; if they are these marks, then they are parallel.1222

If it is one time, then all of the lines with one will be parallel.1226

For these, all of the lines with two will be parallel; if you have another pair of parallel lines, then you can do those three times.1233

OK, the last example: we are going to do a proof.1244

Write a two-column proof: before we begin, we should always look at what is given, what you have to prove, and the diagram.1248

Look at it and see how you are going to get from point A to point B.1260

This is what we are trying to prove: that AB is parallel to EF.1268

Angle 1 and angle 2 are congruent--this is congruent; and then, angle 1 is also congruent to 3.1273

So, one time is congruent...two times are congruent.1299

We know that, since these two angles are congruent, and these two angles are congruent,1310

I know that, since these two angles are going to have some kind of special relationship,1317

my theorem and my postulate say that, if they have a special relationship, then the lines are parallel.1323

So, I am going to just do this step-by-step: here are my statements; my reasons I will put right here.1329

Statement #1: We know that we have to write the given, so it is that (let me write it a little bit higher; I am out of room)...1345

#1 is that angle 1 is congruent to angle 2, and angle 1 is congruent to angle 3.1353

And the reason for this is that it is given.1364

#2: Well, if angle 1 is congruent to angle 2, and angle 1 is congruent to angle 3, then I can say that angle 2 is congruent to angle 3.1370

So, I will read it this way: If angle 2 is congruent to angle 1, and angle 1 is congruent to angle 3, then angle 2 is congruent to angle 3.1387

And this is the transitive property of congruency--not equality, but congruency.1396

#1: If a is equal to b, and b = c, then a = c; that is the transitive property, and we are using congruency, so it is not equality; it is congruency.1412

Then, from there, since I proved that angle 2 is congruent to angle 3, I know that they are alternate interior angles.1427

Alternate interior angles are congruent; that means that I can just say that, since alternate interior angles are congruent, then these lines are parallel.1445

Step 3: You can say that angles 2 and 3 are labeled as alternate interior angles.1456

Or, you can just go ahead and write out what the "prove" statement is--what you are trying to prove,1471

since you already proved that they are congruent, and they are alternate interior angles.1476

Depending on how your teacher wants you to set this up, this will be either step 3 or step 4.1481

My reason is going to be: If alternate interior angles are congruent, then the lines...1491

now, this is not the complete theorem, but you can just shorten it...are parallel.1506

And that would be the proof; and make sure (again, since we haven't done proofs in a while)1517

that the given statement always comes first, and the "prove" statement always comes last.1523

It is like you are trying to get from point A to point B.1526

If you are driving somewhere--you start from your house, and you are driving to school--your house is point A, and your school is point B.1530

There are steps to get there; it is the same thing--proofs are exactly the same way.1541

You need to have your steps to get from point A to point B.1545

That is it for this lesson; we will see you soon; thanks for watching Educator.com.1550

Welcome back to Educator.com.0000

This next lesson is on parallel lines and distance.0002

The first one: the distance between a point and a line: we have a point here, and we have a line.0010

And to find the distance between them, you have to find the length of the segment that is perpendicular.0017

If you were to get a ruler, you need to find the distance--how far away this point is from this line.0026

You are going to take your ruler, and you are going to make it so that the ruler is going to be perpendicular to the line.0035

So, imagine if that point is you; you are that point; you are standing in the room, and this line is the wall in front of you.0043

So, if you have to find the distance between you (this point right here) and the wall (the line), then you don't find the distance this way.0056

You don't go diagonally to the wall; if you are facing directly on the wall, you have to measure the distance0067

so that your pathway, that length of that segment, is going to be perpendicular to that wall.0074

If you are going to find the distance, you are not going to do this; that is not the distance.0090

It has to be from the point to the line so that it is perpendicular, so this right here will be the distance.0098

Now, to find the distance between two parallel lines: we know that parallel lines never intersect.0112

They are always going to have the same distance between them, no matter where you measure it from.0119

Now, that word right there is called "equidistant"; that means that, no matter where you measure the distance, it is always going to be the same.0126

And that is equidistant--an "equal distance," so "equidistant."0138

And so then, that means that, if you want to find the distance between them, you can just find it for any point on the line,0145

and make sure, just like we just went over, that the point and the line are perpendicular.0155

This segment right here: to find the distance, they have to be perpendicular to each other.0163

This would be the distance; if you look for it right here, as long as it is perpendicular, that will also be the distance.0168

That is the distance between two parallel lines.0179

Let's go over a few examples: Draw the segment that represents the distance from point A to EF.0187

And this is segment EF; so here is point A; here is segment EF.0195

I want to draw the segment that represents the distance.0205

If I go like this, the distance this way to the line, that is going to be incorrect, because that is not perpendicular.0212

This right here, what I drew (it is supposed to be a segment)--that is not perpendicular.0225

What you have to do: you have to extend this out; make sure that it is lined up, extend it out...and then you would draw that.0231

This is your distance, and the same thing here: we are going to extend this out, and then draw the segment that is perpendicular to it.0247

And you measure it, and that is the distance.0262

Let's do a couple more: here is point A; here are two parallel lines and point A; you want to find the distance from this point to this line.0267

So, now, if I go like this, is that the distance? No, this is not the distance.0277

You have to make sure to draw it like this, so that it is perpendicular; and that would be the distance.0288

Here is point A; here is segment EF; in this case, you would draw it; that is perpendicular.0297

Graph the given equation (there is your equation) and plot the point; construct a perpendicular segment, and find the distance from the point to the line.0315

So, first, let's draw the line; y = 2x - 1: this is y = mx + b; b is your y-intercept; that means I have to go to -1 and plot that as my y-intercept.0324

Here is my x; here is my y; and then, my slope is 2/1--remember, this is rise over run.0344

If my rise is 2, and it is positive 2, that means that I am going to go up 2, and then I am going to move to the right 1, because that is a positive.0356

For positive, you go right; for negative, you go left.0367

And then, you are going to go again: 1, 2, and you can just keep doing that.0370

You could go from this point--go to -2 and -1, and that is still going to give you a positive 2.0373

There is my line, right there; and then, the point is 0, 1, 2, 3.0381

First, we have graphed the equation; we have plotted the point; construct a perpendicular segment, and find the distance.0395

If we know that this right here, the slope of this line, was 2/1, that means that0407

if I draw a perpendicular segment from this point, the slope is going to be the negative reciprocal.0419

Remember: if they are perpendicular lines, then the slope is going to be the negative reciprocal.0429

For this other line, the perpendicular line, my slope will be -1/2.0438

Now, this can either mean -1 over 2, or 1 over -2; it is the same thing.0447

So, if I make it -1 over 2, from here, -1 is my rise, over run; so for the rise, since it is -1, I am going to go down 1, and 1, 2--over 2.0454

Down 1, and over 2, down 1 and over 2...and then, it is going to look something like this.0471

That is your perpendicular segment.0492

Now, to find the distance from that point to this point on the line, first of all, this is an intersection point between these two lines;0493

so, how can I find the intersection point between these two lines?0503

Well, I have my equation here; this is my first equation; my second equation, right here, is just going to be y =...0509

my slope was -1/2x; my y-intercept was 3; so all I did was to plug in my slope and my y-intercept.0525

The slope is -1/2; the y-intercept is 3; and then, you can just solve those two out, so y = 2x - 1.0536

I am going to use the substitution method...plus 3...equals 2x - 1.0547

And then, if I subtract the 2x over to this side, this is going to be -5/2x is equal to...0555

I am going to subtract the 3 over there, so I am going to get a 4.0566

Now, all I am doing is solving this out; remember, this is systems.0569

And then, to solve for x, I need to multiply this whole thing by the reciprocal, so -2/5.0575

That way, this is going to all cancel and give me 1; multiply that side by it; so x is going to equal 8/5--there is my x, and then my y.0584

I just need to plug it back into one of these; it doesn't matter which one; let's plug it into this one.0606

2 times x is 16/5 - 1; so this will be y = 2(8/5) - 1; here is 16/5 - 1, or I can make this 5/5, and that is going to be 11/5.0610

So, my point is this point right here; it is 8/5 and 11/5.0633

Now, the whole point of doing that (let me just explain), of finding this point, is so that you can find the distance between the two points.0650

If you have this point, and you have this point, you can use the distance formula to find how long the segment is.0662

All of this work was just to find this point right here, because that is the point where they intersected.0671

So then, I am going to use those two points; I am trying to find the distance between (0,3) and (8/5,11/5).0679

The distance formula is the square root of (x2 - x1)2 + (y2 - y1)2.0700

The distance is going to be (0 - 8/5)2 + (3 - 11/5)2.0712

This is -8/5; that is 64/25, plus...3 - 11/5; if you want to subtract these, you need a common denominator.0733

My common denominator is going to be 5; that is 15/5 - 11/5.0747

So, this is going to be 4/5, and that is squared; so it is going to be 16/25.0755

And then, the square root of...I have a common denominator, so this is going to be...80/25 (let me make some room over here):0775

that is the same thing as the square root of 80 over the square root of 25.0799

The square root of 80: this is 8 times 10; 8 is 4 times 2...let me just show you a quick way;0803

I know this is a little off of this lesson, but to simplify square roots, if you have √80, and you want to simplify it,0820

then you can just do the factor tree: this is 8 and 10; this is 4 and 2 (circle it if it is prime), 2 and 2;0832

whenever you have a pair of the same number, that comes out, so this is going to be 2.0846

Here is another 2 that is common; now, this one doesn't have a partner, doesn't have another 5 to come out of the radical.0855

So, only when they have a partner, only when there is two of the same number, can they come out.0870

2 comes out right there; then these 2's come out right there.0876

Now, the 5 has to stay in the radical; and then, you just multiply these two numbers.0881

And then, the square root of 25, we know, is 5; so this is going to be 4√5/5, and that is the distance.0888

Now, I know that this seems like a lot of work, but all we did was0897

construct a perpendicular segment by making the slope the negative reciprocal of this line.0904

Then, we found this point right here, where these two lines intersect; it has to be perpendicular.0915

So then, using this point and that point, we found the distance; and that is it--this is the distance between these two points.0923

All right, the next example: Find the distance between the two parallel lines.0938

OK, now, here I know that it has to be perpendicular wherever I want to find the distance,0943

as long as the segment that is made from the points between them has to be perpendicular.0958

Let's say I am going to use this point right here; then this is going to be perpendicular--that is going to be my distance.0965

Now, this point, I know is (0,2); this point, I know, is going to be 1 and 1/2...0978

1.5, or maybe 3/2 (the same thing): 3/2, and right here, that is going to be 1/2.0992

And I know that, because it is halfway between these two.0999

And the same here: the slope is 1, so everything is going to be half.1002

You can also, if you want, look at it this way: if I continue this on, then it is going to be halfway between this point right here1009

and this point right here, in the same way that this is halfway between this point and this point.1018

That is how I know that it is half.1026

This point is 3/2, 1, and 1/2; so now I have to find the distance between those two points.1028

The distance formula, again, is (x2 - x1), or (x1 - x2), squared, + (y2 - y1)2.1039

I find the square root again; I am going to subtract the x's: 0 - 3/2, squared, plus 2 - 1/2, squared.1057

Then, this is going to be (3/2)2; (3/2)2 is 9/4, plus...2 - 1/2 is 3/2, or 1 and 1/2,1073

and I know that because 2 becomes 4/2, minus 1/2 is 3/2; that squared is 9/4.1090

The square root of...there is already a common denominator, so that is 18/4...(make sure this goes all the way down),1104

which is √18/√4; for √18, you don't have to do the factor tree,1115

because you know that 9, a perfect square, is a factor of 18; it is going to be 3√2/2.1122

All you do is make sure that you just have the two points, whenever you try to find the distance between a point and a line or two parallel lines.1136

Then make sure that you have the two points, so that the segment that connects them is going to be perpendicular to the lines.1145

And then, you just use the distance formula; it is (0 - 3/2)2 + (2 - 1/2)2.1152

And then, that way, you get (3/2)2, which is 9/4, and then this, which is 3/2, squared, is going to be 9/4 again.1163

√18 over √4 simplifies to 3√2/2.1173

That is it for this lesson; thank you for watching Educator.com.1184

Welcome back to Educator.com.0000

In this next lesson, we are going to talk about triangles and some different ways we can classify them.0003

A triangle is a three-sided polygon; now, a polygon is any figure that has sides and is closed.0011

A polygon can look like that, as long as it is closed, and all of the sides are straight.0027

If I have that, it would not be considered a polygon; if I have just maybe something like this,0036

that is not a polygon, either, because it is not closed--that is not an example of a polygon.0047

A triangle is a polygon with three sides; now, the triangle is made up of sides, vertices, and angles.0056

The sides are these right here, side AB, side BC, and side CA (or AC).0066

The vertices are the points; the word "vertices" is the plural of "vertex," so if I just talk about one, then I would just say "vertex."0083

But since I have three, it is "vertices"; and that would just be point A, point B, and point C--those are my vertices.0098

The angles: now, for this angle right here, let's say, I can say "angle ABC," or I can just say "angle B,"0116

because as long as there is only one angle...if I have a line that is coming out through here,0127

then I will have several different angles, so I can't label it angle B;0134

but as long as there is only one single angle from that vertex, then you can label the angle by that point.0137

This angle right here can be called angle B; this angle can be called angle A, since there is no other angle there.0147

This angle can be called angle C, or you can just do it the other way: angle ABC, angle BCA, angle BAC.0157

But this is the easiest way: angle B, angle A, and angle C; that is the triangle.0170

Now, to classify triangles, we can classify them in two ways: by their angles and by their sides.0181

These are the ways that we can classify the triangles by their angles, meaning that, based on the angles of the triangle, we have different names for them.0191

The first one: an acute triangle...well, we know that an acute angle is an angle that measures less than 90, smaller than 90.0200

So, an acute triangle is a triangle where all of the angles are acute; all of the angles measure less than 90.0209

The next one: obtuse triangle: one angle is obtuse.0226

Now, all triangles have at least two acute angles.0231

The way we classify these other ones: for this one, an obtuse angle is an angle that measures greater than 90;0243

only one of them will be obtuse--there is no way that you can have a triangle with two obtuse angles.0253

That means two of the angles (since we have three) are going to be acute, and then one of them is going to be obtuse.0260

But all triangles must have at least two acute angles; so here is an obtuse triangle; here is your obtuse angle, and then your acute angles.0268

The same thing for the next one, a right triangle: a right triangle is when only one angle is a right angle, like this.0285

Again, the other two angles must be acute.0300

Now, if I try to draw two angles of a triangle to be right, see how there is no way that that could be a triangle,0304

because a triangle, remember, has to have three sides; "tri" in triangle means three.0322

Now, an equiangular triangle means that all of the angles are equal--"equal angle"--can you see the two words formed in there?0340

So, equiangular is when all of the angles are congruent.0352

Now, if all of the angles are congruent, then each angle is going to measure 60 degrees.0357

Now, we are going to go over this later on; but the three angles of a triangle have to add up to 180.0368

If all three angles are congruent, then I just do 180 divided by 3, because each angle has to have the same number of degrees.0378

So, 180 divided by 3 is going to be 60; so this will be 60, 60, and then 60.0390

And that is an equiangular triangle; and of course, an equiangular triangle is an acute triangle, because 60 degrees is an acute angle; all three angles are acute.0396

In an equiangular triangle, all angles are congruent; they all measure 60 degrees.0412

OK, so then, classifying triangles by size: we just went over by angles, depending on how the angles look--0419

if it is a right angle in the triangle, an obtuse angle, or an acute angle.0426

We can also classify triangles by the size.0433

If I have a triangle where no two sides are congruent--all three sides are different lengths (like this is 3, 4, 5), then this is a scalene triangle.0438

So, I can show this by making little slash marks; I can make this once; then this, I will do twice to show that these are not congruent;0456

and then I will do this one three times--that shows that none of these sides are congruent to each other.0465

And that is a scalene triangle.0473

The next one, an isosceles triangle, is when I have at least two sides being congruent.0477

And then, an equilateral triangle is when all of the sides are congruent.0493

So, an equilateral triangle will be considered an isosceles triangle, because any time you have at least0499

(meaning two or three) sides being congruent, then it is considered isosceles.0505

But this is the more specific name if all three are congruent.0510

So, you would call this an equilateral triangle; but it is also considered an isosceles triangle.0514

Now, remember this one: equilateral; equiangular is when all of the angles are congruent, and equilateral is when all of the sides are congruent.0523

These are some ways you can classify triangles by the sides.0532

Isosceles triangle: I see here that these two sides are congruent, which is an isosceles triangle.0540

If I label this as triangle ABC, then I can say that triangle ABC is isosceles.0549

Any time you have a triangle, you can label it like this, triangle ABC, just like when you have an angle, you can say angle A.0560

So, if you have a triangle, you are going to say triangle ABC.0566

This right here, this side that is not congruent to the other two sides, is called your base.0573

These two sides that are congruent are called your legs.0587

Now, the two angles that are formed from the legs to the base are called base angles.0598

Base angles would be this angle right here and this angle right here.0609

And then, this angle right here, or that point, is called your vertex.0618

This right here, that is formed from the two congruent legs, is called your vertex.0627

So, we have legs; we have a vertex; we have the base; and then, you have base angles.0634

And this makes up your isosceles triangle.0641

And then, just to go over the right triangle: these are also called your legs, but this one is not called the base; it is called a hypotenuse.0644

That is just to review over that.0668

In the isosceles triangle, these two are congruent, so they are called your legs; this one is your base.0672

Now, don't think that the base is the side always on the bottom; no.0679

It depends on the triangle: I can move this triangle around and make it look like this, and I can say that these two sides are congruent.0684

Then, this would be my base, and these are my legs.0694

Find x, AB, BC, and AC; let me just write this to show that they are my segments.0701

To solve this, if I want to solve for x, I know that these two are congruent.0714

Since they are congruent, I can just make these equal to each other.0722

2x + 3 = 3x - 2; then, if I subtract the 3x, I get -x =...I subtract 3 over, so I get -5; x is 5.0726

There is my x; AB is (and I am not going to put a line over this one, because I am finding the lengths;0742

if I am finding the measure, then I don't put the line over it) 2 times 5, plus 3; so that is 10 + 3, is 13.0753

And then, for BC, since I know x, even though this side is the base, and it doesn't have anything to do with these sides;0772

since I know x, I can solve for BC; that is 5 + 1, so BC = 6.0783

And then, AC is 3(5) - 2, so that is 15, minus 2 is 13; and that is everything.0794

So again, if you have an isosceles triangle, and they want you to solve for x, then you can just say that,0811

since these two sides are congruent, you can make these two congruent.0818

All right, let's go over a few examples now: Classify each triangle with the given angle measures by its angles and sides.0826

We have to classify each of these triangles with the given measures in two ways: by its angles and by its sides.0835

The first one: the angles are here, and the sides here.0845

To classify this triangle by its angles, look: I have one that is really big--that is an obtuse angle; that means that this would be an obtuse triangle.0859

I am just going to draw a little triangle right there.0875

And then, by its sides, remember: sides are scalene, isosceles, and equilateral.0877

No two sides are congruent (the same), so this would be a scalene triangle.0886

The next one: 50, 50, and 80: by its angles--look at the angle measures: they are all acute angles.0897

So then, this triangle would be an acute triangle.0905

And then, with its sides, are any of them the same?0912

We have two that are the same; these two are the same; then, I have an isosceles triangle.0915

The next one: this is an acute triangle, because they are all the same; but more specifically, this would be an equiangular triangle.0926

And then, since they are all the same, even though it could be isosceles (because isosceles is two or more), a more specific name would be equilateral.0946

And the last one: 30, 60, 90: well, I have one angle that is a right angle, so this is a right triangle.0964

And then, my sides: this would be scalene, because they are all different.0975

The next example: We are going to fill in the blanks with "always," "sometimes," or "never."0989

Scalene triangles are [always/sometimes/never] isosceles.0996

Well, we know that a scalene triangle is when they are all different.1001

An isosceles triangle is when we have two or more sides that are the same.1006

So, a scalene triangle would never be isosceles, because in order for the triangle to be scalene, all of the sides have to be different.1011

In order for the triangle to be isosceles, you have to have two or more the same, so this would be "never" isosceles.1022

The next one: Obtuse triangles [always/sometimes/never] have two obtuse angles.1034

Well, if I draw an obtuse angle here, and then I draw an obtuse angle here, this has to be greater than 90,1042

and this has to be greater than 90; a triangle has to have three sides only, so there is no way for that to be a triangle.1050

So, this is "never."1058

Equilateral triangles are [always/sometimes/never] acute triangles.1066

If they are equilateral, that means that it has to be like this.1075

Can we ever have an obtuse triangle where they are the same?--no, because you know that this can only be extra long.1082

And then, if we have a right triangle, no, because we can't have the hypotenuse be the same length as one of the legs.1092

So, equilateral triangles are "always" acute.1099

Isosceles triangles are [always/sometimes/never] equilateral triangles.1110

An isosceles triangle is like this or like this; "isosceles" can be two or three sides being congruent.1115

Now, sometimes they are like this, and sometimes they are like that.1127

That means that sometimes they are going to be equilateral.1131

Because isosceles triangles can be considered equilateral triangles (but not always; if it is only two, then it is not; if it is three, then it is), it is "sometimes."1137

Acute triangles are [always/sometimes/never] equiangular.1151

If I have an acute triangle, yes, it could be equiangular; but can I draw an acute triangle that is not equiangular?1158

How about like this? This is not equiangular, but they are all acute angles.1170

Sometimes it could be like this, or sometimes it could be like this; "sometimes"--acute triangles are sometimes equiangular.1184

If I have, let's say, 50, 50, and 80; see how these are all acute angles.1200

They are acute angles, but then it is not equiangular; only if they are 60, 60, 60 are they equiangular.1215

I can have other acute triangles that are not equiangular.1223

OK, the next example: Find x and all of the sides of the isosceles triangle.1231

Here, it is this side and this side that are congruent.1238

Now again, this is the base (even though it is not at the bottom, it is still called the base).1244

These are my legs, the vertex, and the base angles.1250

I am going to make 9x + 12 equal to 11x - 4, because those sides are congruent.1256

Then I am going to solve this out; I am going to subtract 11x over there, so I get -2x = -16; x = +8--there is my x.1267

Then, I have to look for all of my sides: AB is going to be 9 times 8 plus 12; that is 72 + 12, so AB is equal to 84.1281

AC is 11 times 8, minus 4; that is 88 - 4, which is 84; that is AC.1305

And notice how they are the same; they have to be the same, because that is the whole point.1320

They are isosceles; these are congruent; we make them equal to each other so that they will be the same.1323

And then, BC is 2 times 8 plus 10, so this is 16 + 10, so BC is 26.1329

And that is it for this problem.1347

Use the distance formula to classify the triangle by its sides.1351

Here is my triangle, ABC; and then, you are going to use the distance formula to find what kind of sides there are.1355

So, if I find the distance of A to C, then I will find the length of this side.1364

Then I can use the distance formula to find the length of that side, and again do the same thing here.1372

And then, you are going to compare those distances of the three sides.1376

That means I will have to use the distance formula three times, for each of the sides.1382

There is A, and there is B, and there is C.1386

Now, A (let me just write it out) is at point...here is -2...1, 2, 3; B is at (-1,3); and C is at...here is 1, 2, 3...-1.1389

Let's find AB first; it doesn't matter which one you find first.1426

AB: I am going to use points A and B, these two; this is going to be x2, and then I will just write out the distance formula again right here.1431

All you do is subtract the x's, square that number, and add it to that number.1441

AB is (-2 - -1)2 + (-3 - 3)2; -2...minus a negative is the same thing as plus,1453

so that is going to be -1 squared, is 1, plus -3 - 3, is -6, squared is +36; so this is going to be √37; there is AB.1482

And then, BC is B and C: it is (-1 - 3)2 + (3 - -1)2.1503

And then, -1 - 3 is -4, squared is 16; plus...this is going to be 3 + 1; that is 4; this is 16; and then, that is the square root of 32.1523

This can be simplified; remember from the last lesson: if you want to simplify radicals,1544

square roots, then you have to write this down; you can do a factor tree.1551

This is going to be 16, and this is going to be 2; or you can just do 8 and 4; it doesn't matter.1557

Circle it if it is prime; now, 16 here is a perfect square, so I can just go ahead and do that.1563

But just to show you: it is going to be 8 and 2 (or 4 and 4; it doesn't matter), 4 and 2, 2 and 2.1568

Whenever you have two of the same number, it is going to come out of the radical a single time.1578

So then, as long as there are two of the same number, it comes out as one.1589

So then, that comes out as a 2; these come out as a 2; and then, this one is still left.1595

Whatever is left has to stay inside, and then, whatever came out (2 came out here, and 2 came out here)...that becomes 4√2.1604

And then again, you know that is 16 times 2; we know that because it is 16 + 16; and then, just do that.1616

That is BC; and then, AC is going to be (-2 - 3)2 + (-3 - -1)2.1626

So, this is going to be -5, squared is 25, plus...this is going to be -2, because it is going to be plus; -2 squared is 4; so this is...1647

OK, let me just write it here: AC = √29.1666

And you can't simplify that any more.1675

Now that I finished my distance formula, applying the distance formula to each of the sides,1680

AC was √29; BC is 4√2; and then, AB is √37.1687

Now, see how all three sides are different: the whole point is to classify the triangle by its sides.1696

That means that it is either going to be a scalene, an isosceles, or an equilateral triangle.<