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I have a circle here whose circumference is 18 pi. So if we were to measure all the way around the circle, we would get 18 pi. And we also have a central angle here. So this is the center of the circle. And this central angle that I'm about to draw has a measure of 10 degrees. So this angle right over here is 10 degrees. And what I'm curious about is the length of the arc that subtends that central angle.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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So this is the center of the circle. And this central angle that I'm about to draw has a measure of 10 degrees. So this angle right over here is 10 degrees. And what I'm curious about is the length of the arc that subtends that central angle. So what is the length of what I just did in magenta? And one way to think about it, or actually maybe the way to think about it, is that the ratio of this arc length to the entire circumference, let me write this down, arc length to the entire circumference should be the same as the ratio of the central angle to the total number of angles if you were to go all the way around the circle, so 2, 360 degrees. So let's just think about that.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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And what I'm curious about is the length of the arc that subtends that central angle. So what is the length of what I just did in magenta? And one way to think about it, or actually maybe the way to think about it, is that the ratio of this arc length to the entire circumference, let me write this down, arc length to the entire circumference should be the same as the ratio of the central angle to the total number of angles if you were to go all the way around the circle, so 2, 360 degrees. So let's just think about that. We know the circumference is 18 pi. So we know the circumference is 18 pi. We're looking for the arc length.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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So let's just think about that. We know the circumference is 18 pi. So we know the circumference is 18 pi. We're looking for the arc length. I'm just going to call that a, a for arc length. That's what we're going to try to solve for. We know that the central angle is 10 degrees.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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We're looking for the arc length. I'm just going to call that a, a for arc length. That's what we're going to try to solve for. We know that the central angle is 10 degrees. So you have 10 degrees over 360 degrees. Over 360. So we could simplify this by multiplying both sides by 18 pi.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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We know that the central angle is 10 degrees. So you have 10 degrees over 360 degrees. Over 360. So we could simplify this by multiplying both sides by 18 pi. Multiply both sides by 18 pi. And we get that our arc length is equal to, well, 10 over 360 is the same thing as 1 over 36. So it's equal to 1 over 36 times 18 pi, which is just so.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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So we could simplify this by multiplying both sides by 18 pi. Multiply both sides by 18 pi. And we get that our arc length is equal to, well, 10 over 360 is the same thing as 1 over 36. So it's equal to 1 over 36 times 18 pi, which is just so. It's 18 pi over 36, which is the same thing as pi over 2. So this arc right over here is going to be pi over 2, whatever units we're talking about, long. Now let's think about another scenario.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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So it's equal to 1 over 36 times 18 pi, which is just so. It's 18 pi over 36, which is the same thing as pi over 2. So this arc right over here is going to be pi over 2, whatever units we're talking about, long. Now let's think about another scenario. Let's imagine another, or the same circle. So it's the same circle here. Our circumference is still 18 pi.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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Now let's think about another scenario. Let's imagine another, or the same circle. So it's the same circle here. Our circumference is still 18 pi. Circumference is still 18 pi. There are people having a conference behind me or something. That's why you might hear those mumbling voices.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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Our circumference is still 18 pi. Circumference is still 18 pi. There are people having a conference behind me or something. That's why you might hear those mumbling voices. But this circumference is also 18 pi. So this is also 18 pi. But now I'm going to make the central angle, I'm going to make it an obtuse angle.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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That's why you might hear those mumbling voices. But this circumference is also 18 pi. So this is also 18 pi. But now I'm going to make the central angle, I'm going to make it an obtuse angle. So the central angle now, so let's say we were to start right over here. This is one side of the angle. I'm going to go and make a 350 degree angle.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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But now I'm going to make the central angle, I'm going to make it an obtuse angle. So the central angle now, so let's say we were to start right over here. This is one side of the angle. I'm going to go and make a 350 degree angle. So I'm going to go all the way around like that. So this right over here is a 350 degree angle. And now I'm curious about this arc that subtends this really huge angle.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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I'm going to go and make a 350 degree angle. So I'm going to go all the way around like that. So this right over here is a 350 degree angle. And now I'm curious about this arc that subtends this really huge angle. So now I want to figure out this arc length. So all of this, I want to figure out this arc length, the arc that subtends this really obtuse angle right over here. Well, same exact logic.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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And now I'm curious about this arc that subtends this really huge angle. So now I want to figure out this arc length. So all of this, I want to figure out this arc length, the arc that subtends this really obtuse angle right over here. Well, same exact logic. The ratio between our arc length, the ratio between our arc length, a, and the circumference of the entire circle, 18 pi, should be the same as the ratio between our central angle, our central angle that the arc subtends, so 350, over the total number of degrees in a circle, over 360. So multiply both sides by 18 pi, we get a is equal to, let's see, this is 35 times 18 over 36 pi. So 350 divided by 360 is 35 over 36.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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Well, same exact logic. The ratio between our arc length, the ratio between our arc length, a, and the circumference of the entire circle, 18 pi, should be the same as the ratio between our central angle, our central angle that the arc subtends, so 350, over the total number of degrees in a circle, over 360. So multiply both sides by 18 pi, we get a is equal to, let's see, this is 35 times 18 over 36 pi. So 350 divided by 360 is 35 over 36. So this is 35 times 18 times pi over 36. Well, both 36 and 18 are divisible by 18, so let's divide them both by 18. And so we are left with 35 over 2 pi, or we could even say, let me just write it that way, 35 pi over 2, or if you wanted to write it as a decimal, this would be 17.5 pi.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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So 350 divided by 360 is 35 over 36. So this is 35 times 18 times pi over 36. Well, both 36 and 18 are divisible by 18, so let's divide them both by 18. And so we are left with 35 over 2 pi, or we could even say, let me just write it that way, 35 pi over 2, or if you wanted to write it as a decimal, this would be 17.5 pi. Now, does this make sense? This right over here, this other arc length, when our central angle was 10 degrees, this had an arc length of 0.5 pi. So when you add these two together, this arc length and this arc length, 0.5 plus 17.5, you get to 18 pi, which was the circumference, which makes complete sense, because if you add these angles, 10 degrees and 350 degrees, you get 360 degrees in a circle.
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Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3
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And we have our little tool here on Khan Academy where we can construct a quadrilateral. And we need to construct the reflection of triangle ABCD. And so what we can do is, let me scroll down a little bit so we can see the entire coordinate axis. We wanna find the reflection across the x-axis. So I'm gonna reflect point by point. And actually, let me just move this whole thing down here so that we can see what is going on a little bit clearer. So let's just first reflect point, let me move this a little bit out of the way.
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Example reflecting quadrilateral over x axis.mp3
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We wanna find the reflection across the x-axis. So I'm gonna reflect point by point. And actually, let me just move this whole thing down here so that we can see what is going on a little bit clearer. So let's just first reflect point, let me move this a little bit out of the way. So let's first reflect point A. So we're gonna reflect across the x-axis. A is four units above the x-axis.
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Example reflecting quadrilateral over x axis.mp3
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So let's just first reflect point, let me move this a little bit out of the way. So let's first reflect point A. So we're gonna reflect across the x-axis. A is four units above the x-axis. One, two, three, four. So its image, A prime we could say, would be four units below the x-axis. So one, two, three, four.
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Example reflecting quadrilateral over x axis.mp3
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A is four units above the x-axis. One, two, three, four. So its image, A prime we could say, would be four units below the x-axis. So one, two, three, four. So let's make this right over here A, A prime. I'm having trouble putting the, see if I move these other characters around. Okay, there you go.
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Example reflecting quadrilateral over x axis.mp3
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So one, two, three, four. So let's make this right over here A, A prime. I'm having trouble putting the, see if I move these other characters around. Okay, there you go. So this is gonna be my A prime. Now let me try B. B is two units above the x-axis.
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Example reflecting quadrilateral over x axis.mp3
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Okay, there you go. So this is gonna be my A prime. Now let me try B. B is two units above the x-axis. So B prime is gonna have the same x-coordinate, but it's gonna be two units below the x-axis. So let's make this our B. So this is our B right over here.
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Example reflecting quadrilateral over x axis.mp3
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B is two units above the x-axis. So B prime is gonna have the same x-coordinate, but it's gonna be two units below the x-axis. So let's make this our B. So this is our B right over here. Now let's make this our C. C right here has the x-coordinate of negative five and a y-coordinate of negative four. Now C prime would have the same x-coordinate, but instead of being four units below the x-axis, it'll be four units above the x-axis. So we would have the coordinates negative five comma positive four.
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Example reflecting quadrilateral over x axis.mp3
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So this is our B right over here. Now let's make this our C. C right here has the x-coordinate of negative five and a y-coordinate of negative four. Now C prime would have the same x-coordinate, but instead of being four units below the x-axis, it'll be four units above the x-axis. So we would have the coordinates negative five comma positive four. So this is gonna be our C here. So this goes to negative five. One, two, three, positive four.
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Example reflecting quadrilateral over x axis.mp3
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So we would have the coordinates negative five comma positive four. So this is gonna be our C here. So this goes to negative five. One, two, three, positive four. And then last but not least, D. And so let's see, D right now is at negative two comma negative one. If we were to reflect across the x-axis, instead of being one unit below the x-axis, we'll be one unit above the x-axis, and we'll keep our x-coordinate of negative two. And so there you have it.
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Example reflecting quadrilateral over x axis.mp3
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One, two, three, positive four. And then last but not least, D. And so let's see, D right now is at negative two comma negative one. If we were to reflect across the x-axis, instead of being one unit below the x-axis, we'll be one unit above the x-axis, and we'll keep our x-coordinate of negative two. And so there you have it. We have constructed the reflection of ABCD across the x-axis. And what's interesting about this example is that the original quadrilateral is on top of the x-axis. So you can kinda see this top part of the quadrilateral gets reflected below it, and this bottom part of the quadrilateral gets reflected above it.
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Example reflecting quadrilateral over x axis.mp3
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And I also touched a little bit on the idea of an axiom or postulate, but I want to be clear. Sometimes you will hear this referred to as a side, side, side theorem, and sometimes you'll hear it as a side, side, side postulate or axiom. Postulate or axiom. And I think it's worth differentiating what these mean. A postulate or an axiom is something that you just assume. You assume from the get-go, while a theorem is something you prove using more basic or using some postulates or axioms. So, in really, in all of mathematics, you make some core assumptions.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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And I think it's worth differentiating what these mean. A postulate or an axiom is something that you just assume. You assume from the get-go, while a theorem is something you prove using more basic or using some postulates or axioms. So, in really, in all of mathematics, you make some core assumptions. You make some core assumptions. You call these the axioms or the postulates. And then using those, you try to prove theorems.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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So, in really, in all of mathematics, you make some core assumptions. You make some core assumptions. You call these the axioms or the postulates. And then using those, you try to prove theorems. So maybe using that one, I can prove some theorem over here. And maybe using that theorem and then this axiom, I can prove another theorem over here. And then using both of those theorems, I can prove another theorem over here.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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And then using those, you try to prove theorems. So maybe using that one, I can prove some theorem over here. And maybe using that theorem and then this axiom, I can prove another theorem over here. And then using both of those theorems, I can prove another theorem over here. I think you get the picture. So this axiom might lead us to this theorem, and these two might lead us to this theorem right over here. And we essentially try to build our knowledge or we build a mathematics around these core assumptions.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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And then using both of those theorems, I can prove another theorem over here. I think you get the picture. So this axiom might lead us to this theorem, and these two might lead us to this theorem right over here. And we essentially try to build our knowledge or we build a mathematics around these core assumptions. In an introductory geometry class, we kind of, we don't rigorously prove side, side, side. We don't rigorously prove the side, side, side theorem. And that's why in a lot of geometry classes, you kind of just take it as a given, as a postulate or an axiom.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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And we essentially try to build our knowledge or we build a mathematics around these core assumptions. In an introductory geometry class, we kind of, we don't rigorously prove side, side, side. We don't rigorously prove the side, side, side theorem. And that's why in a lot of geometry classes, you kind of just take it as a given, as a postulate or an axiom. And the whole reason why I'm doing this is one, just so you know the difference between the words theorem and postulate or axiom, and also so that you don't get confused. It is just a given, but in a lot of books, and I've looked at several books, they do refer to it as a side, side, side theorem, even though they never prove it rigorously. They do just assume it.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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And that's why in a lot of geometry classes, you kind of just take it as a given, as a postulate or an axiom. And the whole reason why I'm doing this is one, just so you know the difference between the words theorem and postulate or axiom, and also so that you don't get confused. It is just a given, but in a lot of books, and I've looked at several books, they do refer to it as a side, side, side theorem, even though they never prove it rigorously. They do just assume it. So it really is more of a postulate or an axiom. Now with that out of the way, we're just going to assume going forward that we just know that this is true. We're going to take it as a given.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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They do just assume it. So it really is more of a postulate or an axiom. Now with that out of the way, we're just going to assume going forward that we just know that this is true. We're going to take it as a given. I want to show you that we can already do something pretty useful with it. So let's say that we have a circle. There's many useful things that we can already do with it.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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We're going to take it as a given. I want to show you that we can already do something pretty useful with it. So let's say that we have a circle. There's many useful things that we can already do with it. And this circle has a center right here at A. And let's say that we have a chord in this circle that is not a diameter. So let me draw a chord here.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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There's many useful things that we can already do with it. And this circle has a center right here at A. And let's say that we have a chord in this circle that is not a diameter. So let me draw a chord here. So let me draw a chord in this circle. So it's a kind of a segment of a secant line. And let's say that I have a line that bisects this chord from the center.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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So let me draw a chord here. So let me draw a chord in this circle. So it's a kind of a segment of a secant line. And let's say that I have a line that bisects this chord from the center. I guess I could call it a radius because I'm going to go from the center to the edge of the circle right over there. So I'm going to the center to the circle itself. And when I say bisects it, so these are all, I'm just setting up the problem right now.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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And let's say that I have a line that bisects this chord from the center. I guess I could call it a radius because I'm going to go from the center to the edge of the circle right over there. So I'm going to the center to the circle itself. And when I say bisects it, so these are all, I'm just setting up the problem right now. When I say bisecting it, it means it splits that line segment in half. So what it tells us is that the length of this segment right over here is going to be equivalent to the length of this segment right over there. And what I want to do is, so this is, I've set it up, I have a circle.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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And when I say bisects it, so these are all, I'm just setting up the problem right now. When I say bisecting it, it means it splits that line segment in half. So what it tells us is that the length of this segment right over here is going to be equivalent to the length of this segment right over there. And what I want to do is, so this is, I've set it up, I have a circle. This radius bisects this chord right over here. And what I want to do is prove, the goal here is to prove that it bisects this chord at a right angle. Or another way to say it, let me add some points here.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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And what I want to do is, so this is, I've set it up, I have a circle. This radius bisects this chord right over here. And what I want to do is prove, the goal here is to prove that it bisects this chord at a right angle. Or another way to say it, let me add some points here. Let's call this B, let's call this C, and let's call this D. I want to prove that segment AB is perpendicular, it intersects it at a right angle, is perpendicular to segment CD. And as you can imagine, I'm going to prove it pretty much using the side, side, side, whatever you want to call it, side, side, side theorem, postulate, or axiom. So let's do it, let's think about it this way.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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Or another way to say it, let me add some points here. Let's call this B, let's call this C, and let's call this D. I want to prove that segment AB is perpendicular, it intersects it at a right angle, is perpendicular to segment CD. And as you can imagine, I'm going to prove it pretty much using the side, side, side, whatever you want to call it, side, side, side theorem, postulate, or axiom. So let's do it, let's think about it this way. So you can imagine, if I'm going to use this, I need to have some triangles. There's no triangles here right now. But I can construct triangles, and I can construct triangles based on things I know.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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So let's do it, let's think about it this way. So you can imagine, if I'm going to use this, I need to have some triangles. There's no triangles here right now. But I can construct triangles, and I can construct triangles based on things I know. For example, I can construct, this has some radius, so let's call this, that's a radius right over here. The length of that is just going to be the radius of the circle. But I can also do it right over here.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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But I can construct triangles, and I can construct triangles based on things I know. For example, I can construct, this has some radius, so let's call this, that's a radius right over here. The length of that is just going to be the radius of the circle. But I can also do it right over here. The length of AC is also going to be the radius of the circle. So we know that these two lines have the same length, which is the radius of the circle. We could say that AD is congruent to AC, or they have the exact same lengths.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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But I can also do it right over here. The length of AC is also going to be the radius of the circle. So we know that these two lines have the same length, which is the radius of the circle. We could say that AD is congruent to AC, or they have the exact same lengths. We know from the setup in the problem, that this segment is equal in length to this segment over here. We could even, let me add a point here so I can refer to it. So if I call that point E, we know from the setup in the problem that CE is congruent to ED, or they have the same length.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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We could say that AD is congruent to AC, or they have the exact same lengths. We know from the setup in the problem, that this segment is equal in length to this segment over here. We could even, let me add a point here so I can refer to it. So if I call that point E, we know from the setup in the problem that CE is congruent to ED, or they have the same length. CE has the same length as ED. We also know that both of these triangles, the one here on the left and the one here on the right, they both share the side EA. So EA is clearly equal to EA.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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So if I call that point E, we know from the setup in the problem that CE is congruent to ED, or they have the same length. CE has the same length as ED. We also know that both of these triangles, the one here on the left and the one here on the right, they both share the side EA. So EA is clearly equal to EA. So this is clearly equal to itself. It's the same side. The same side is being used for both triangles.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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So EA is clearly equal to EA. So this is clearly equal to itself. It's the same side. The same side is being used for both triangles. The triangles are adjacent to each other. So we see a situation where we have two different triangles that have corresponding sides being equal. This side is equivalent to this side right over here.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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The same side is being used for both triangles. The triangles are adjacent to each other. So we see a situation where we have two different triangles that have corresponding sides being equal. This side is equivalent to this side right over here. This side is equal in length to that side over there. And then we have, obviously AE is equivalent to itself. It's a side on both of them.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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This side is equivalent to this side right over here. This side is equal in length to that side over there. And then we have, obviously AE is equivalent to itself. It's a side on both of them. It's the corresponding side on both of these triangles. So by side, side, side, we know that triangle ABC is congruent to triangle AE. Sorry, it's not ABC.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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It's a side on both of them. It's the corresponding side on both of these triangles. So by side, side, side, we know that triangle ABC is congruent to triangle AE. Sorry, it's not ABC. It's AEC. We know, actually let me write it over here. By side, side, side, we know that triangle AEC is congruent to triangle AED.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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Sorry, it's not ABC. It's AEC. We know, actually let me write it over here. By side, side, side, we know that triangle AEC is congruent to triangle AED. But how does that help us? How does that help us knowing that, you know, we used our little theorem, but how does that actually help us here? Well, what's cool is once we know that two triangles are congruent, so because they are congruent, that tells us, so from that we can deduce that all the angles are the same.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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By side, side, side, we know that triangle AEC is congruent to triangle AED. But how does that help us? How does that help us knowing that, you know, we used our little theorem, but how does that actually help us here? Well, what's cool is once we know that two triangles are congruent, so because they are congruent, that tells us, so from that we can deduce that all the angles are the same. And in particular, we can deduce that this angle right over here, that the measure of angle CEA is equivalent to the measure of angle DEA. And the reason why that's useful is that we also see just by looking at this that they are supplementary to each other. Their adjacent angles, their outer sides form a straight angle.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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Well, what's cool is once we know that two triangles are congruent, so because they are congruent, that tells us, so from that we can deduce that all the angles are the same. And in particular, we can deduce that this angle right over here, that the measure of angle CEA is equivalent to the measure of angle DEA. And the reason why that's useful is that we also see just by looking at this that they are supplementary to each other. Their adjacent angles, their outer sides form a straight angle. So CEA is supplementary and equivalent to DEA. So they're also supplementary. So we also have the measure of angle CEA plus the measure of angle DEA is equal to 180 degrees.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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Their adjacent angles, their outer sides form a straight angle. So CEA is supplementary and equivalent to DEA. So they're also supplementary. So we also have the measure of angle CEA plus the measure of angle DEA is equal to 180 degrees. But they're equivalent to each other. So I could replace the measure of DEA with the measure of CEA. Measure of angle CEA.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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So we also have the measure of angle CEA plus the measure of angle DEA is equal to 180 degrees. But they're equivalent to each other. So I could replace the measure of DEA with the measure of CEA. Measure of angle CEA. Or I could rewrite this as 2 times the measure of angle CEA is equal to 180 degrees. Or I could divide both sides by 2, and I say the measure of angle CEA is equal to 90 degrees, which is going to be the same as the measure of angle DEA because they're equivalent. So we know that this angle right over here is 90 degrees, so I can do it with that little box.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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Measure of angle CEA. Or I could rewrite this as 2 times the measure of angle CEA is equal to 180 degrees. Or I could divide both sides by 2, and I say the measure of angle CEA is equal to 90 degrees, which is going to be the same as the measure of angle DEA because they're equivalent. So we know that this angle right over here is 90 degrees, so I can do it with that little box. And this angle right over here is 90 degrees. And because AB intersects where it intersects CD, we have a 90-degree angle here and there, and we could also prove that it's over there as well. They are perpendicular to each other.
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SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3
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So let's just start by drawing ourselves a pyramid, and I'll draw one with a rectangular base, but depending on how we look at the formula, we could have a more general version. But a pyramid looks something like this. And you might get a sense of what the formula for the volume of a pyramid might be. If we say this dimension right over here is x, this dimension right over here, the length right over here is y, and then you have a height of this pyramid. If you were to go from the center straight to the top, or if you were to measure this distance right over here, which is the height of the pyramid, you just call that, let's call that z. And so you might say, well, I'm dealing with three dimensions, so maybe I multiply the three dimensions together, and that would give you volume in terms of units, but if you just multiplied xy times z, that would give you the volume of the entire rectangular prism that contains the pyramid. So that would give you the volume of this thing, which is clearly bigger, has a larger volume than the pyramid itself.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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If we say this dimension right over here is x, this dimension right over here, the length right over here is y, and then you have a height of this pyramid. If you were to go from the center straight to the top, or if you were to measure this distance right over here, which is the height of the pyramid, you just call that, let's call that z. And so you might say, well, I'm dealing with three dimensions, so maybe I multiply the three dimensions together, and that would give you volume in terms of units, but if you just multiplied xy times z, that would give you the volume of the entire rectangular prism that contains the pyramid. So that would give you the volume of this thing, which is clearly bigger, has a larger volume than the pyramid itself. The pyramid is fully contained inside of it, so this would be the tip of the pyramid on the surface, just like that. And so you might get a sense that, all right, maybe the volume of the pyramid is equal to x times y times z times some constant. And what we're going to do in this video is have an argument as to what that constant should be, assuming that this, the volume of the pyramid is roughly of this structure.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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So that would give you the volume of this thing, which is clearly bigger, has a larger volume than the pyramid itself. The pyramid is fully contained inside of it, so this would be the tip of the pyramid on the surface, just like that. And so you might get a sense that, all right, maybe the volume of the pyramid is equal to x times y times z times some constant. And what we're going to do in this video is have an argument as to what that constant should be, assuming that this, the volume of the pyramid is roughly of this structure. And to help us with that, let's draw a larger rectangular prism and break it up into six pyramids that completely make up the volume of the rectangular prism. So first, let's imagine a pyramid that looks something like this, where its width is x, its depth is y, so that could be its base, and its height is halfway up the rectangular prism. So if the rectangular prism has height z, the pyramid's height is going to be z over two.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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And what we're going to do in this video is have an argument as to what that constant should be, assuming that this, the volume of the pyramid is roughly of this structure. And to help us with that, let's draw a larger rectangular prism and break it up into six pyramids that completely make up the volume of the rectangular prism. So first, let's imagine a pyramid that looks something like this, where its width is x, its depth is y, so that could be its base, and its height is halfway up the rectangular prism. So if the rectangular prism has height z, the pyramid's height is going to be z over two. Now, what would be the volume of that pyramid based on what we just saw over here? Well, that volume would be equal to some constant k times x times y, not times z, times the height of the pyramid, times z over two. So it'd be x times y times z over two.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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So if the rectangular prism has height z, the pyramid's height is going to be z over two. Now, what would be the volume of that pyramid based on what we just saw over here? Well, that volume would be equal to some constant k times x times y, not times z, times the height of the pyramid, times z over two. So it'd be x times y times z over two. I'll just write times z over two. Or actually, we could even write it this way, x, y, z over two. Now, I can construct another pyramid that has the exact same dimensions.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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So it'd be x times y times z over two. I'll just write times z over two. Or actually, we could even write it this way, x, y, z over two. Now, I can construct another pyramid that has the exact same dimensions. If I were to just flip that existing pyramid on its head and look something like this, this pyramid also has dimensions of an x-width, a y-depth, and a z over two height. So its volume would be this as well. Now, what is the combined volume of these two pyramids?
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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Now, I can construct another pyramid that has the exact same dimensions. If I were to just flip that existing pyramid on its head and look something like this, this pyramid also has dimensions of an x-width, a y-depth, and a z over two height. So its volume would be this as well. Now, what is the combined volume of these two pyramids? Well, it's just going to be this times two. So the combined volume of these pyramids, let me just draw it that way. So these two pyramids that look something like this, I'm gonna try to color-code it.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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Now, what is the combined volume of these two pyramids? Well, it's just going to be this times two. So the combined volume of these pyramids, let me just draw it that way. So these two pyramids that look something like this, I'm gonna try to color-code it. We have two of them, so two times their volume is going to be equal to, well, two times this is just going to be k times x, y, z. K, x, y, and z. And we have more pyramids to deal with. For example, I have this pyramid right over here where this face is its base.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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So these two pyramids that look something like this, I'm gonna try to color-code it. We have two of them, so two times their volume is going to be equal to, well, two times this is just going to be k times x, y, z. K, x, y, and z. And we have more pyramids to deal with. For example, I have this pyramid right over here where this face is its base. And then if I try to draw the pyramid, it looks something like this, this one right over there. Now, what is its volume going to be? Its volume is going to be equal to k times its base is y times z.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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For example, I have this pyramid right over here where this face is its base. And then if I try to draw the pyramid, it looks something like this, this one right over there. Now, what is its volume going to be? Its volume is going to be equal to k times its base is y times z. So k, y, z. And what's its height? Well, its height is going to be half of x.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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Its volume is going to be equal to k times its base is y times z. So k, y, z. And what's its height? Well, its height is going to be half of x. So this height right over here is half of x. So it's k times y times z times x over two, or I could say times x, and then divide everything by two. Now, I have another pyramid that has the exact same dimensions.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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Well, its height is going to be half of x. So this height right over here is half of x. So it's k times y times z times x over two, or I could say times x, and then divide everything by two. Now, I have another pyramid that has the exact same dimensions. This one over here, if I try to draw it on the other face opposite the one we just saw, essentially if we just flip this one over, has the exact same dimensions. So one way to think about it, we have two pyramids that look like that with those types of dimensions. This is for an arbitrary rectangular prism that we are dealing with.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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Now, I have another pyramid that has the exact same dimensions. This one over here, if I try to draw it on the other face opposite the one we just saw, essentially if we just flip this one over, has the exact same dimensions. So one way to think about it, we have two pyramids that look like that with those types of dimensions. This is for an arbitrary rectangular prism that we are dealing with. So I have two of these. And so if you have two of their volumes, what's it going to be? It's going to be two times this expression.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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This is for an arbitrary rectangular prism that we are dealing with. So I have two of these. And so if you have two of their volumes, what's it going to be? It's going to be two times this expression. So it's going to be k times x, y, z. X, y, z. Interesting. And then last but not least, we have two more pyramids.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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It's going to be two times this expression. So it's going to be k times x, y, z. X, y, z. Interesting. And then last but not least, we have two more pyramids. We have this one that has a face that has the base right over here. That's its base, and if it was transparent, you'd be able to see where I'm drawing right here. And then you have one on the opposite side, right over there on the other side.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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And then last but not least, we have two more pyramids. We have this one that has a face that has the base right over here. That's its base, and if it was transparent, you'd be able to see where I'm drawing right here. And then you have one on the opposite side, right over there on the other side. And I could just say if you were to flip this around. And so by the exact same argument, so let me just draw it. So we have two of these, two of these pyramids.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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And then you have one on the opposite side, right over there on the other side. And I could just say if you were to flip this around. And so by the exact same argument, so let me just draw it. So we have two of these, two of these pyramids. My best to draw it. So times two. So each of them would have a volume of what?
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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So we have two of these, two of these pyramids. My best to draw it. So times two. So each of them would have a volume of what? Each of them, their base is x times z. So it's going to be k times x times z. That's the area of their base.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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So each of them would have a volume of what? Each of them, their base is x times z. So it's going to be k times x times z. That's the area of their base. And then what is their height? Well, each of them has a height of y over two. So times y over two.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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That's the area of their base. And then what is their height? Well, each of them has a height of y over two. So times y over two. And I have two of those pyramids, so I'm gonna multiply those by two. The twos cancel out, so I'm just left with k times x, y, z. So k times x, y, z.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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So times y over two. And I have two of those pyramids, so I'm gonna multiply those by two. The twos cancel out, so I'm just left with k times x, y, z. So k times x, y, z. Now, one of the interesting things that we've just stumbled on in this is seeing that even though these pyramids have different dimensions and look different, they all have actually the same volume, which is interesting in and of themselves. And so if we were to add up the volumes of all of the pyramids here and use this formula to express them, so if I were to add all of them together, that should be equal to the volume of the entire rectangular prism. And then maybe we can figure out k. So the volume of the entire rectangular prism is x, y, z, x times y times z.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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So k times x, y, z. Now, one of the interesting things that we've just stumbled on in this is seeing that even though these pyramids have different dimensions and look different, they all have actually the same volume, which is interesting in and of themselves. And so if we were to add up the volumes of all of the pyramids here and use this formula to express them, so if I were to add all of them together, that should be equal to the volume of the entire rectangular prism. And then maybe we can figure out k. So the volume of the entire rectangular prism is x, y, z, x times y times z. And then that's gotta be equal to the sum of these. So that's going to be equal to k x, y, z plus k x, y, z plus k x, y, z. Or you could say that's going to be equal to three k x, y, z.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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And then maybe we can figure out k. So the volume of the entire rectangular prism is x, y, z, x times y times z. And then that's gotta be equal to the sum of these. So that's going to be equal to k x, y, z plus k x, y, z plus k x, y, z. Or you could say that's going to be equal to three k x, y, z. All I did is I said, let me just add up the volume from all of these pyramids. And so what do we get for k? Well, we could divide both sides by three x, y, z to solve for k. Three x, y, z, three x, y, z.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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Or you could say that's going to be equal to three k x, y, z. All I did is I said, let me just add up the volume from all of these pyramids. And so what do we get for k? Well, we could divide both sides by three x, y, z to solve for k. Three x, y, z, three x, y, z. And we are left with, on the right-hand side, everything cancels out, we're just left with a k. And on the left-hand side, we're left with a 1 3rd. And so we get k is equal to 1 3rd. K is equal to 1 3rd.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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Well, we could divide both sides by three x, y, z to solve for k. Three x, y, z, three x, y, z. And we are left with, on the right-hand side, everything cancels out, we're just left with a k. And on the left-hand side, we're left with a 1 3rd. And so we get k is equal to 1 3rd. K is equal to 1 3rd. And there you have it, that's our argument for why the volume of a pyramid is 1 3rd times the dimensions of the base times the height. So you might see it written that way, or you might see it written as 1 3rd times base. And so if x times y is the base, so the area of the base, so the base area, times the height, which in this case is z.
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Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3
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What we're going to do in this video is learn to construct congruent angles. And we're going to do it with, of course, a pen or a pencil here. I'm going to use a ruler as a straight edge. And then I'm going to use a tool known as a compass, which looks a little bit fancy. But what it allows us to do, and we'll apply using it in a little bit, is it allows us to draw perfect circles or arcs of a given radius. You pivot on one point here, and then you use your pen or your pencil to trace out the arc or the circle. So let's just start with this angle right over here.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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And then I'm going to use a tool known as a compass, which looks a little bit fancy. But what it allows us to do, and we'll apply using it in a little bit, is it allows us to draw perfect circles or arcs of a given radius. You pivot on one point here, and then you use your pen or your pencil to trace out the arc or the circle. So let's just start with this angle right over here. And I'm going to construct an angle that is congruent to it. So let me make the vertex of my second angle right over there. And then let me draw one of the rays that originates at that vertex.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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So let's just start with this angle right over here. And I'm going to construct an angle that is congruent to it. So let me make the vertex of my second angle right over there. And then let me draw one of the rays that originates at that vertex. And I'm gonna put this angle in a different orientation just to show that they don't even have to have the same orientation. So it's going to look something like that. That's one of the rays.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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And then let me draw one of the rays that originates at that vertex. And I'm gonna put this angle in a different orientation just to show that they don't even have to have the same orientation. So it's going to look something like that. That's one of the rays. But then we have to figure out where do we put, where do we put the other ray so that the two angles are congruent? And this is where our compass is going to be really useful. So what I'm going to do is put the pivot point of a compass, of the compass, right at the vertex of the first angle.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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That's one of the rays. But then we have to figure out where do we put, where do we put the other ray so that the two angles are congruent? And this is where our compass is going to be really useful. So what I'm going to do is put the pivot point of a compass, of the compass, right at the vertex of the first angle. And I'm going to draw out an arc like this. And what's useful about the compass is you can keep the radius constant. And you can see it intersects our first two rays at points, let's just call this B and C. And I could call this point A right over here.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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So what I'm going to do is put the pivot point of a compass, of the compass, right at the vertex of the first angle. And I'm going to draw out an arc like this. And what's useful about the compass is you can keep the radius constant. And you can see it intersects our first two rays at points, let's just call this B and C. And I could call this point A right over here. And so let me, now that I have my compass with the exact right radius right now, let me draw that right over here. But this alone won't allow us to draw the angle just yet. But let me draw it like this.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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And you can see it intersects our first two rays at points, let's just call this B and C. And I could call this point A right over here. And so let me, now that I have my compass with the exact right radius right now, let me draw that right over here. But this alone won't allow us to draw the angle just yet. But let me draw it like this. And that is pretty good. And let's call this point right over here D, I don't know, I'll call this one E. And I want to figure out where to put my third point F so I can define ray EF so that these two angles are congruent. And what I can do is take my compass again and get a clear sense of the distance between C and B by adjusting my compass.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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But let me draw it like this. And that is pretty good. And let's call this point right over here D, I don't know, I'll call this one E. And I want to figure out where to put my third point F so I can define ray EF so that these two angles are congruent. And what I can do is take my compass again and get a clear sense of the distance between C and B by adjusting my compass. So one point is on C and my pencil is on B. So I have, let me get this right, so I have this distance right over here. I know this distance.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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And what I can do is take my compass again and get a clear sense of the distance between C and B by adjusting my compass. So one point is on C and my pencil is on B. So I have, let me get this right, so I have this distance right over here. I know this distance. And I have adjusted my compass accordingly. So I can get that same distance right over there. And so you can now imagine where I'm going to draw that second ray.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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I know this distance. And I have adjusted my compass accordingly. So I can get that same distance right over there. And so you can now imagine where I'm going to draw that second ray. That second ray, if I put point F right over here, my second ray I can just draw between, starting at point E right over here, going through point F. I could draw that a little bit neater. So it would look like that, my second ray. Ignore that first little line I drew, I'm using a pen, which I don't recommend for you to do it.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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And so you can now imagine where I'm going to draw that second ray. That second ray, if I put point F right over here, my second ray I can just draw between, starting at point E right over here, going through point F. I could draw that a little bit neater. So it would look like that, my second ray. Ignore that first little line I drew, I'm using a pen, which I don't recommend for you to do it. I'm doing it so that you can see it on this video. Now how do we know that this angle is now congruent to this angle right over here? Well, one way to do it is to think about triangle BAC, triangle BAC, and triangle, let's just call it DFE.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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Ignore that first little line I drew, I'm using a pen, which I don't recommend for you to do it. I'm doing it so that you can see it on this video. Now how do we know that this angle is now congruent to this angle right over here? Well, one way to do it is to think about triangle BAC, triangle BAC, and triangle, let's just call it DFE. So this triangle right over here. When we drew that first arc, we know that the distance between AC is equivalent to the distance between AB, and we kept the compass radius the same. So we know that's also the distance between EF and the distance between ED.
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Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3
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So given this diagram, we need to figure out what the length of CF right over here is. And you might already guess that this will have to do something with similar triangles. Because at least it looks that triangle CFE is similar to ABE. And the intuition there is it's kind of embedded inside of it. And we're going to prove that to ourselves. And it also looks like triangle CFB is going to be similar to triangle DEB. But once again, we're going to have to prove that to ourselves.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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And the intuition there is it's kind of embedded inside of it. And we're going to prove that to ourselves. And it also looks like triangle CFB is going to be similar to triangle DEB. But once again, we're going to have to prove that to ourselves. And then maybe we can deal with all the ratios of the different sides to CF right over here. And then actually figure out what CF is going to be. So first, let's prove to ourselves that these definitely are similar triangles.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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But once again, we're going to have to prove that to ourselves. And then maybe we can deal with all the ratios of the different sides to CF right over here. And then actually figure out what CF is going to be. So first, let's prove to ourselves that these definitely are similar triangles. So you have this 90 degree angle in ABE. And you have this 90 degree angle in CFE. If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So first, let's prove to ourselves that these definitely are similar triangles. So you have this 90 degree angle in ABE. And you have this 90 degree angle in CFE. If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar. And we can either show that, look, they both share this angle right over here. Angle CEF is the same as angle AEB. So we've shown two angles, two corresponding angles in these triangles.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar. And we can either show that, look, they both share this angle right over here. Angle CEF is the same as angle AEB. So we've shown two angles, two corresponding angles in these triangles. This is an angle in both triangles. They are congruent, so the triangles are going to be similar. You could also show that this line is parallel to this line, because obviously these two angles are the same.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So we've shown two angles, two corresponding angles in these triangles. This is an angle in both triangles. They are congruent, so the triangles are going to be similar. You could also show that this line is parallel to this line, because obviously these two angles are the same. And so these angles will also be the same. So they're definitely similar triangles. So let's just write that down, get that out of the way.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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You could also show that this line is parallel to this line, because obviously these two angles are the same. And so these angles will also be the same. So they're definitely similar triangles. So let's just write that down, get that out of the way. We know that triangle ABE is similar to triangle CFE. And you want to make sure you get in the right order. F is where the 90 degree angle is.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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