WEBVTT mathematics/geometry/pyo
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Welcome back to Educator.com.
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This lesson is on measuring segments; let's begin.
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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.
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But if we are just talking about this part from point A to point B, that is a segment,
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where we are not talking about all of this--just between point A and point B.
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That would be a segment, and we call that segment AB.
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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.
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But for the segments, we are just writing a line like that--a segment above it.
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A segment is like a line with two endpoints.
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And if we are talking about the measure of AB, the measure of AB is like the distance between A and B.
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And when you are talking about the measure, you don't write the bar over it--you just leave it as AB.
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So, when you are just talking about the segment itself, then you would put the bar over it;
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if not, then you are just leaving it as AB; OK.
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Ruler Postulate: this has to do with the distance: The points on any line can be paired with real numbers
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so that, given any two points, P and Q, on the line, P corresponds to 0, and Q corresponds to a positive number,
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just like when you want to measure something--you use a ruler, and you put the 0 at the first point;
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and then you see how long whatever you are trying to measure is.
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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--
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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.
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That is what I am saying: the first number corresponds to 0.
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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.
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And when you use the ruler postulate, you can also find the distance of two points on the number line, using absolute value.
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I can just subtract these two numbers, 2 minus 8; but I am going to use absolute value.
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So then, 2 - 8 is -6; the absolute value of it is going to make it 6.
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Remember: absolute value is the distance from 0; so if it is -6, how far away is -6 from 0? 6, right?
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So, absolute value just makes everything positive.
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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.
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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.
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That is also 6; so either way, your answer is going to be 6.
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Another example: Find the distance between this point, -4, and...let's see...5.
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I am going to find the distance from -4 to +5.
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I can do the same thing: the absolute value of -4 - 5; this is the absolute value of -9, which is 9.
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From -4 to 5...they are 9 units apart from each other.
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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.
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You don't have to do it twice; I am just trying to show you that you will have the same distance,
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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.
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That is the Ruler Postulate.
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The next one, the Segment Addition Postulate: you will use this postulate many, many, many times throughout the course.
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A postulate, to review, is a math statement that is assumed to be true.
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Unlike theorems...theorems are also math statements, but theorems have to be proved
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in order for us to use them, to accept it as true; but postulates we can just assume to be true.
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So, any time there is a postulate, then we don't have to question its value or its truth.
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We can just assume that it is true, and then just go ahead and use it.
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Segment Addition Postulate: if Q is between P and R, then QP + PR = PR.
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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.
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QP + QR = PR: so Q is between P and R--let me just write that out here.
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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.
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It is...if I have a part of something, and I have another part of something, it makes up the whole thing.
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And if PQ + QR equals PR, then Q is between P and R; so you can use it both ways.
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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.
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Or if I give you that this is 10, and then the whole thing is 15, then this is going to be 5, right?
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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...
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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--
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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.
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That is the Segment Addition Postulate.
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Find BC if B is between A and C and AB is 2x - 4; BC is 3x - 1; and AC equals 14.
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Find BC if B is between A and C...let's draw that out: here is A and C, and B is between them.
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It doesn't have to be in the middle, just anywhere in between those two points.
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If I have B right here, then we know that AB, this segment, plus this segment, equals the whole segment, AC.
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So, AB is 2x - 4; and BC is 3x - 1; the whole thing, AC, is 14.
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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?
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So, I am going to do 2x - 4, that segment, plus 3x - 1, equals 14.
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So, here, to solve this, 2x + 3x is 5x; and then, -4 - 1 is -5; that equals 14.
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If I add 5 to that, 5x = 19, and x = 19/5; OK.
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And they want you to find BC; now, you found x, but always look to see what they are asking for.
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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,
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because if I am going to subtract these two fractions, then I need a common denominator.
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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.
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And then, this will be 52/5; you could just leave it as a fraction.
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And notice one thing: how these BC's, these segments, don't have the bars over them.
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And that is because you are dealing with measure: whenever you have a segment equaling its value,
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equaling some number, some distance, some value, then you are not going to have the bar over it, because you are talking about measure.
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The next one: Write a mathematical sentence given segments ED and EF.
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This is using the Segment Addition Postulate; that is the kind of mathematical sentence it wants you to give.
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ED and EF--that is all you are given, segments ED and EF.
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Well, here is E; there is an E in both; that means that E has to be in the middle.
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E has to be here, because since it is in both segments, that is the only way I can have E in both.
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So then, here, this will be D, and this can be F.
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Now, this can be F, and this can be D; it doesn't really matter,
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as long as you have E in the middle, somewhere in between, and then D and F as the endpoints of the whole segment.
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And to use the Segment Addition Postulate, I can say that DE or ED, plus EF, equals DF; we just write it like that.
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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.
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You probably remember this from algebra: if you have a right triangle...
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now, you have to keep that in mind; the Pythagorean Theorem can only be used on right triangles;
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a right triangle, and you use it to find a missing side.
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You have to be given two out of the three sides--any two of the three sides--to find the missing side.
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That is what you use the Pythagorean Theorem for--only for right triangles, though.
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So, a² + b² = c²: that is the formula.
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You have to make the hypotenuse c; this has to be c.
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Now, just to go over, briefly, the Pythagorean Theorem, we have a right triangle, again.
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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.
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The hypotenuse will always be c; it doesn't have to be c, but from the formula, whatever you make this equal to--
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the square of the sum of the two sides--has to be...
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I'm sorry: you have to square each side, and then you take the sum of that; it equals the hypotenuse squared.
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OK, and let me just go over this part right here.
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If we have this side as 3, then you square it, and it becomes 9.
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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.
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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.
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And then, it just means that, when you add up the two, it is going to be the area of this square right here.
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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.
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And then, that just makes this side 5.
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a²...back to this formula...+ b² = c²: you just have to square the side,
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square the other side, add them up, and then you get the hypotenuse squared.
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Let's do a problem: Find the missing side.
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I have the legs, the measure of the two legs, and I need to find the hypotenuse.
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So, that is 4 squared, plus 3 squared, equals the hypotenuse squared; so I can just call that c squared.
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So, 4 squared is 16, plus 9, equals c squared; 25 = c².
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And then, from here, I need to square root both; so this is going to become 5.
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Now, normally, when you square root something, you are going to have a plus/minus that number;
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but since we are dealing with distance, the measure of the side, it has to be positive; so this right here is 5.
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OK, the distance formula: The distance between any two points with coordinates (x₁,y₁)
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and (x₂,y₂), is given by the formula d = the square root
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of the difference of the x's, squared, plus the difference of the y's, squared.
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Here, this distance formula is used to find the distance between two points.
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And we know that a point is (x,y); and the reason why it is labeled like this...
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you have to be careful; I have seen students use these numbers as exponents.
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Instead of writing it like that, they would say (x²,y²); that is not true.
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This is just saying that it is the first x and the first y; so this is from the first point.
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They are saying, "OK, well, this is (x,y) of the first point; and this is the second point."
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And that is all that these little numbers are saying; they are saying the first x and first y,
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from the first point, and the second x and the second y from the second point.
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x₂ just means the second x, the x in the second point.
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And it doesn't matter which one you make the first point, and which one you make the second point;
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just whichever point you decide to make first and second, then you just keep that as x₂, x₁, y₂, and y₁.
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Find DE for this point and point D and E; so then, I can make this (x₁,y₁), my first point;
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and then this would be (x₂,y₂)--not (x²,y²); it is (x₂,y₂), the second point.
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Then, the distance between these two points...I take my second x (that is 1) minus the other x, so minus -6, squared,
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plus the second y, 5, minus the other y (minus 2), squared.
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1 - -6: minus negative is the same thing as plus the whole thing, so this will be 7 squared plus 3 squared.
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7 squared is 49; plus 3 squared is 9; this is going to be 58.
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Now, 58--from here, you would have to simplify it.
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To see if you can simplify it, the easiest way to simplify square roots--you can just do the factor tree.
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I just want to do this quickly, just to show you.
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A factor of 58 is going to be 2...2 and 29.
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Now, 2 is a prime number, so I am going to circle that.
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And then, 29: do we have any factors of 29? No, we don't.
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So, this will be the answer; we know that we can't simplify it.
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The distance between these two points is going to be the square root of 58.
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Let's do a few examples that have to do with the whole lesson.
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Find each measure: AC: here is A, and here is C.
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You can use the Ruler Postulate, and you can make this point correspond to 0.
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And then, you see what C will become, what number C will correspond to.
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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;
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so that is going to be the absolute value of -8, which becomes 8.
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BE: absolute value...where is B? -1, minus E (is 9)...so this is the absolute value of -10, which is 10.
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And then, DC: the absolute value of...D is 5, minus 2.
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Now, see how I went backwards, because that was DC.
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It doesn't matter: you can do CD or DC; with segments, you can go either way.
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So, DC is 5 - 2 or 2 - 5; it is going to be the absolute value of 3, which is 3.
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The next example: Given that U is between T and V, find the missing measure.
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Here, let's see: there is T; there is V; and then, U is just anywhere in between.
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TU, this right here, is 4; TV, the whole thing, is 11.
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So, if the whole thing is 11, and this is 4, well, I know that this plus this is the whole thing, right?
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So, you can do this two ways: you can make UV become x; I can make TU;
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or plus...UV is x...equals the whole thing, which is 11.
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You can solve it that way, or you can just do the whole thing, minus this segment.
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If you have the whole thing, and you subtract this, then you will get UV.
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You can do it that way, too; if you subtract the 4, you get 7, so UV is 7.
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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.
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So, I know that 3.5 + 6.2 is going to give me TV.
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If I add this up, I get 9.7 = TV.
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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.
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I know that VU, this one, plus TU--these are the parts, and this is the whole thing.
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The whole thing, 5x, equals the sum of its parts, 4x - 1 plus (that is the first part; the second part is) 2x - 1.
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I am just going to add them up; so this will be 6x - 1 - 1...that is -2.
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And then, if I subtract this over, this is going to be -x = -2, which makes x 2.
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Now, look what they are asking for, though: you are not done here.
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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.
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TU is going to be 2 times 2 minus 1, which is 4 minus 1, which is 3; so TU is 3.
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The next example: You are finding the distance between the two points.
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The first one has A at (6,-1) and B at (-8,0); so again, label this as (x₁,y₁);
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this has to be x; this has to be y; you are just labeling as the first x, first y;
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this is also (x,y), but you are labeling it as x₂, the second x, and then the second y.
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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.
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x₂ is -8, minus 6, squared, plus...and then y₂ is 0, minus -1, squared; this is -14 squared, plus 1 squared.
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-14 squared is 196, plus 1...and then this is just going to be the square root of 197.
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And then, the next one: I have two points here: I have A at this point, and I have B at this point.
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Now, even though it is not like this problem, where they give you the coordinates, they are showing you the coordinates.
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They graphed it for you; so then, you have to find the coordinates of the points first.
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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.
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So then, to find the distance between those...let's do it right here:
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it is going to be...you can label this, again: this is (x₁,y₁), (x₂,y₂).
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So, it is -2, the second x, minus the first x, minus 2, squared, plus the second y, -2, minus 1, squared.
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And then, let me continue it right here, so that I have more room to go across:
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the square root of...-2 - 2 is -4, squared; and then plus...this is -3, squared;
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-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.
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The distance between these two points, A and B, is 5.
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And you could just say 5 units.
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The last example: we are going to use the Pythagorean Theorem to find the missing links.
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It has a typo...find...
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And these are both right triangles; I'll just show that...OK.
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The first one: the Pythagorean Theorem is a² + b² = c².
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Here, I am missing this side; I am going to just call that, let's say, b.
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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.
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a² is 5²; plus b², equals 13².
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5 squared is 25, plus b squared, equals 169; and then, subtract the 25; so b² = 144.
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Then, b equals 12, because you square root that; so this is 12.
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The second one: I am going to label this c, because it is the hypotenuse.
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Then: a² + b² = c², so 6² + 8² = c².
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6² is 36, plus 64, equals c²; these together make 100, so c is 10.
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OK, well, that is it for this lesson; thank you for watching Educator.com.