*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 numbers*0093

*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 proved*0321

*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, a*^{2} + b^{2} = c^{2}: 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 a*^{2} + b^{2} = c^{2}, 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

*a*^{2}...back to this formula...+ b^{2} = c^{2}: 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 = c*^{2}.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 (x*_{1},y_{1})1010

*and (x*_{2},y_{2}), 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 (x*^{2},y^{2}); 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

*x*_{2} 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 x*_{2}, x_{1}, y_{2}, and y_{1}.1100

*Find DE for this point and point D and E; so then, I can make this (x*_{1},y_{1}), my first point;1108

*and then this would be (x*_{2},y_{2})--not (x^{2},y^{2}); it is (x_{2},y_{2}), 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 (x*_{1},y_{1});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 x*_{2}, the second x, and then the second y.1559

*The distance formula is the square root of x*_{2}, the second x, minus the other x, squared, plus the difference of the y's, squared.1565

*x*_{2} is -8, minus 6, squared, plus...and then y_{2} 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 (x*_{1},y_{1}), (x_{2},y_{2}).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 a*^{2} + b^{2} = c^{2}.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

*a*^{2} is 5^{2}; plus b^{2}, equals 13^{2}.1816

*5 squared is 25, plus b squared, equals 169; and then, subtract the 25; so b*^{2} = 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: a*^{2} + b^{2} = c^{2}, so 6^{2} + 8^{2} = c^{2}.1857

*6*^{2} is 36, plus 64, equals c^{2}; 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 (x*_{1},y_{1}) and (x_{2},y_{2}),0184

*you are going to use this formula right here: remember from the last lesson: we also used (x*_{1},y_{1})0193

*and (x*_{2},y_{2}) for the distance formula--remember that, for this one,0200

*it is not (x*^{2},y^{2}), 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, (x*_{2},y_{2}): 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 (x*_{1},y_{1}) and (x_{2},y_{2}).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 (x*_{1},y_{1}) and which is one is (x_{2},y_{2}).0429

*So then, we could just make this (x*_{1},y_{1}), and this could be (x_{2},y_{2}); 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: x*_{2} + x_{1}, or x_{1} + x_{2}, 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 x*_{1} + x_{2}; 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 (x*_{1},y_{1})...then what is this going to be? (x_{2},y_{2}), right?0617

*It is as if our coordinates for B are going to be the x*_{2} (this is what we are solving for)...0633

*we are solving for x*_{2}, and we are solving for y_{2}.0640

*Those are the two points that we need.*0644

*Going back to this--well, we know what x*_{1} was, and we know that this whole thing equals this whole thing.0649

*So, I am going to make this a 0, plus x*_{2}/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 x*_{2}, you get 4; so x_{2} = 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 + y*_{2}, divided by 2, equals -1.0703

*And this is what I am solving for, again; so what is y*_{2}? y_{2} 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 + x*_{2} = 2.0759

*And then, the sum of the y's, divided by 2, is -1; so you do 0 + y*_{2}, 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 know*1385

*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, x*_{1} + x_{2}, 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 (x*_{1},y_{1}), or (x_{2},y_{2}).1872

*And then, when you write it out, it is going to be -5 + x*_{2} over 2, and then 2 + y_{2}, 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 + x*_{2}, over 2.1927

*Now, this is not x*^{2}; be careful of that.1933

*That equals -2; this is going to be -5 + x*_{2} equals...you multiply the 2 over to the other side...-4.1936

*You add the 5 over; so x*_{2} = 1.1948

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

*You multiply the 2 over; 2 + y*_{2} = -2; y_{2} = -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 AB*0552

*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 ray*0575

*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 PQR*0748

*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 write*1467

*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 angle*0726

*from each side or vertex...then if this right here is 1, this is 2, this is 3, this is 4, and this is 5*0742

*(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 then*1303

*(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 together*1453

*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 area*0151

*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 triangle*0393

*(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 (y*_{2} - y_{1})/(x_{2} - x_{1}).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 10*0951

*(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 number*1016

*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 base*0103

*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 complicated*0178

*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 bases*0975

*(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 polygon*0008

*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 create*0255

*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 apothem*0413

*(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 πr*^{2}.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 triangle*0695

*(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 squared*0946

*(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 πr*^{2}--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 πr*^{2}; 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 diagram*2124

*(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 side*0151

*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 a*^{2}:b^{2}.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 a*^{2}:b^{2}.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 a*^{2}:b^{2}.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 factor*0252

*of the areas between this one and this one is going to be 2*^{2}:3^{2}, 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 a*^{2}:b^{2}.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 a*^{2}:b^{2}; be careful not to multiply it by 2--you have to square it.0424

*So, 4*^{2} is 16; and 7^{2} 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 3*^{2} to 2^{2}; that is 9:4.0469

*And the third one: they give us the ratio of the areas, so since this is a*^{2} to b^{2}, 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 13*^{2} is 169.0497

*And 144...the square root of that is 12; 12*^{2} = 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 3*^{2}, which is 9, and 1^{2}, 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 areas*0621

*(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 a*^{2} to b^{2}; 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 a*^{2} to b^{2},0911

*which is a*^{2} to...b is 8, so 8^{2}.0929

*Now, that is the same thing as a*^{2}/8^{2}; so we are going to use this ratio and make it equal to these areas.0938

*So, a*^{2} is the same thing as, here, 218, over 166; so the ratio of this area to that area is a^{2}: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) a*^{2} equals 218 times 8^{2} (is 64).0975

*So, from here, you can just divide this 166; a*^{2} =...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 a*^{2} to b^{2}, 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 AC*0018

*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 A*0191

*(this rectangle is region A) is chosen at random, then the probability that the point is in region B*0199

*(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 angle*0264

*(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 is*0429

*(the area of this whole circle is going to be) πr*^{2}.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 πr*^{2}.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 πr*^{2}.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 to*0828

*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 πr*^{2}.0850

*My radius is 4, squared; so then, I can go ahead and cross-multiply.*0859

*360A = 72(π)(4*^{2}); 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 4*^{2}...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 probability*0933

*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 πr*^{2}.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 πr*^{2}, 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 πr*^{2}, 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 a*^{2} + b^{2} = c^{2}.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 πr*^{2}.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 call*0497

*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 base*0768

*(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, ED*0905

*(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 sides*0754

*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 be*0958

*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 πr*^{2}, so it is 2 times πr^{2}.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, h*^{2} + 2.5^{2} = 6^{2}; 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 πr*^{2}; 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 πr*^{2}.0751

*So, it is πrl + πr*^{2}.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...l*^{2} + 6^{2} = 10^{2}.1031

*This is l*^{2} + 36 = 100; then l^{2} 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, 8*^{2} + 6^{2} = the slant height, squared; as you can see, that is the right triangle.1254

*This is 64 + 36 = l*^{2}; 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 πr*^{2}; 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: πr*^{2} 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) r*^{2}, 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 4*^{2}, 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, a*^{2} + b^{2} = c^{2}; 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 30*0950

*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 10*1196

*(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 πr*^{2}.0076

*This is 1/3 times πr*^{2} times the height: 1/3 times π times 3^{2} 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 pyramid*0297

*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; 1/3πr*^{2} 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, πr*^{2},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 25π*0500

*(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 volume*0641

*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 a*^{2} + 2.5^{2} = 11^{2}.0709

*And this is the Pythagorean theorem: a*^{2} + b^{2} equals hypotenuse squared.0719

*Just use your calculator, and get a*^{2}...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 a*^{2} + b^{2} = c^{2}.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 h*^{2} + 11^{2} = 16^{2}, 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 together*1027

*(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 center*0014

*(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 4πr*^{2}.0206

*This is the radius; it is just the area of the circle, times 4; it is 4πr*^{2}, 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, 4πr*^{2}.0237

*And then, to find the volume of the sphere, it will be 4/3πr*^{3}.0249

*The surface area is 4πr*^{2}, and the volume is 4/3πr^{3}.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 4πr*^{2}, 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 4/3πr*^{3}, 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, 4/3πr*^{3};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, πr*^{2}, 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 ahead*0714

*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 volume*0742

*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 congruent*0072

*(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 a*^{2}:b^{2},0139

*and the ratio of the volumes is going to be a*^{3}:b^{3}.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 2*^{2}:3^{2}, 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 2*^{3}:3^{3}; 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...h*^{2} + 24^{2} 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 a*^{2}:b^{2}.0649

*And then, for volume, it is going to be a*^{3}:b^{3}.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 2*^{3} to...what is that one, 3?...3^{3}, 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 that*0820

*(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 (x*_{2} - x_{1})^{2} + (y_{2} - y_{1})^{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 4*^{2} + 2^{2}; this is 16 + 4, is 20; √20...0781

*And then, AC and DF...DF is (5 - 1)*^{2} + (1 + 1)^{2}, so this is 4^{2} + 2^{2},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 image*0170

*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-coordinate*0144

*(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 line*0094

*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 image*0249

*--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 that*0422

*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 going*0719

*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 case*1458

*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 reasoning*0057

*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 here*0885

*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 ABC*1121

*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 line*0148

*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 plane*0506

*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 segments*0163

*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 drawing*0064

*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 same*1196

*(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 out*1323

*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 2*0335

*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 up*0526

*(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 assume*1169

*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 parts*1733

*(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 lines*0319

*(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 *p*0338

*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 lines*1542

*(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 lines*2202

*(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 theorems*0198

*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 transversal*0219

*(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 angles*0286

*(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 angles*0331

*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 lines*0596

*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 lines*0936

*(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 two*1564

*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 line*2276

*(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 (x*_{1},y_{1})0044

*and (x*_{2},y_{2}) is given by this formula right here.0057

*Now, (x*_{1},y_{1}), these numbers right here, and (x_{2},y_{2}), 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 (y*_{2} - y_{1})/(x_{2} - x_{1}).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 x*_{2} - x_{1} 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 y*_{2} - y_{1}, then it has to be x_{2} - x_{1} for the denominator.0168

*It can't be (y*_{2} - y_{1})/(x_{1} - x_{2}).0174

*If you do y*_{2} - y_{1} over here, you cannot do x_{1} - x_{2}.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 x*_{1} and x_{2}, these numbers, can't equal each other,0190

*because if they do, then this denominator is going to become 0.*0197

*If x*_{1} is 5, and x_{2} 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 (y*_{2} - y_{1})/(x_{2} - x_{1});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 (x*_{1},y_{1}), (x_{2},y_{2}), then my slope is going to be (-2 - 3)...0285

*so then, this value is (y*_{2} - y_{1}, 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 (y*_{2} - y_{1})...(-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 1/2*0970

*(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, y*_{2} - y_{1}, over x_{2} - x_{1}.1070

*And that is -6/-3, so we have 2.*1093

*And then, the slope of CD is y*_{2}, -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), (y*_{2} - y_{1}), 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 (y*_{2} - y_{1})/(x_{2} - x_{1}).1338

*Well, I can just fill everything in, except for this, and then use that as x*_{1}.1350

*1/2 is my m; y*_{2}...if this is (x_{1},y_{1}), (x_{2},y_{2}),1356

*y*_{2} 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 just*1646

*(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 transfer*1939

*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 (y*_{2} - y_{1})/(x_{2} - x_{1}),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 (x*_{2},y_{2})--then my slope is y_{2}, which is -5, minus 1, over x_{2}, -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 show*0065

*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 BIK*0604

*(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 relationship*0754

*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 supplementary*0986

*(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 simplified*1160

*(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 distance*0067

*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 that*0407

*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 (x*_{2} - x_{1})^{2} + (y_{2} - y_{1})^{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 was*0897

*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 here*1009

*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 (x*_{2} - x_{1}), or (x_{1} - x_{2}), squared, + (y_{2} - y_{1})^{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 least*0499

*(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 x*_{2}, 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 *