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### Measuring Segments

• Ruler postulate: The points on any line can be paired with real numbers so that, given any two points P and Q on the line, P corresponds to zero, and Q corresponds to a positive number
• If Q is between P and R, then QP + QR = PR
• If PQ + QR = PR, then Q is between P and R
• 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
• Pythagorean Theorem: a2 + b2 = c2
• The distance d between any two points with coordinates (x1, y1) and (x2, y2) is given by the formula:

### Measuring Segments

Find each measure: AE, BD, DE and CE.
• AE = | − 8 − 7| = | − 15| = 15
• BD = | − 2 − 3| = | − 5| = 5
• DE = |3 − 7| = | − 4| = 4
• CE = |0 − 7| = | − 7| = 7
AE = 7; BD = 5; DE = 4; CE = 7
Given that B is between A and C, Find the missing measure.
1. AB = 3, BC = 8, AC = ?
2. AC = 9, BC = 4, AB = ?
• 1. AC = AB + BC = 3 + 8 = 11
• 2. AB = AC − BC = 9 − 4 = 5
1. AC = 11; 2. AB = 5
Given that B is between A and C. Find BC.
1) AB = 5x + 1, BC = x − 2, AC = 17
2) AB = 2, BC = 2x + 2, AC = 3x + 1
• 1) AB + BC = AC
• 5x + 1 + x − 2 = 17
• x = 3
• BC = x − 2 = 3 − 2 = 1
• 2) AB + BC = AC
• 2 + 2x + 2 = 3x + 1
• x = 3
• BC = 2x + 2 = 2 ×3 + 2 = 8
1. BC = 1;
2. BC = 8
Find the distance d between point A( − 1, − 3) and point B (2, 4).
• d = √{(2 − ( − 1))2 + (4 − ( − 3))2} = √{58}
d = √{58}
Find the distance d between points A and B on the coordinate plane in the following figure.
• A(4, 5), B( − 4, − 3)
• d = √{( − 4 − 4)2 + ( − 3 − 5)2} = √{82 + 82} = √{128} = 8√2
d = 8√2
Points A and B are on the coordinate plane, C is between points A and B, AC = 3, find BC.
• A( − 3, 2), B (4, − 3)
• AB = √{(4 − ( − 3))2 + ( − 3 − 2)2} = √{72 + 52} = √{74}
• BC = AB − AC = √{74} − 3
BC = √{74} − 3
For right triangle ABC, ∠ABC = 90o, AB = 6, BC = 5, find AC.
• With Pythagorean Theorem, AC = √{AB2 + BC2}
• AC = √{62 + 52} = √{61}
AC = √{61}
For right triangle ABC, ∠ABC = 90o, AC = 10, AB = 6, find BC.
• With Pythagorean Theorem, BC = √{AC2 − AB2}
• BC = √{102 − 62} = √{100 − 36} = √{64} = 8
BC = 8
For right triangle ABC, ∠ABC = 90o, ∠ADB = 90o, AB = 5, BD = 4, BC = 6, D is between points A and C, find CD.
• With Pythagorean Theorem, AC = √{AB2 + BC2} = √{52 + 62} = √{61}
AD = √{AB2 − BD2} = √{52 − 42} = 3
• CD = AC − AD = √{61} − 3
CD = √{61} − 3
For right triangle ABC, ∠ABC = 90o, ∠ADB = 90o, AD = 3, CD = 5, BD = 4, D is between points A and C, find AB and BC.
• With Pythagorean Theorem, AB = √{AD2 + BD2} = √{32 + 42} = 5
• BC = √{CD2 + BD2} = √{52 + 42} = √{41}
BC = √{41}

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Measuring Segments

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Segments 0:06
• Examples of Segments
• Ruler Postulate 1:30
• Ruler Postulate
• Example and Definition of Segment Addition Postulate
• Example 1: Segment Addition Postulate
• Example 2: Segment Addition Postulate
• Pythagorean Theorem 12:36
• Definition of Pythagorean Theorem
• Pythagorean Theorem, cont. 15:49
• Example: Pythagorean Theorem
• Distance Formula 16:48
• Example and Definition of Distance Formula
• Extra Example 1: Find Each Measure 20:32
• Extra Example 2: Find the Missing Measure 22:11
• Extra Example 3: Find the Distance Between the Two Points 25:36
• Extra Example 4: Pythagorean Theorem 29:33

### Transcription: Measuring Segments

Welcome back to Educator.com.0000

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

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

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

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

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

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

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

A segment is like a line with two endpoints.0057

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

So, absolute value just makes everything positive.0200

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

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

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

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

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

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

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

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

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

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

That is the Ruler Postulate.0299

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

That is the Segment Addition Postulate.0480

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

And then, that just makes this side 5.0925

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

And you could just say 5 units.1768

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

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

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

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

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

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

a2 is 52; plus b2, equals 132.1816

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

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

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

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

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

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