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So plus 11 squared is equal to d squared. So let me just take the calculator out. So the distance, let's just take, if we get 7 squared plus 11 squared is equal to 170. That distance is going to be the square root of that, right? d squared is equal to 170. So let's take the square root of 170 and we get roughly 13.04. So this distance right here, we tried to figure out, is 13.04.
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Distance formula Analytic geometry Geometry Khan Academy.mp3
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Or another way to think about it is, if the shortest side is 1, and I'll do the shortest side, then the medium side, then the longest side. So if the side opposite the 30 degree side is 1, then the side opposite the 60 degree side is square root of 3 times that, so it's going to be square root of 3, and then the hypotenuse is going to be twice that. In the last video, we started with x and we said that the 30 degree side is x over 2, but if the 30 degree side is 1, then this is going to be twice that, so it's going to be 2. This right here is the side opposite the 30 degree side, opposite the 60 degree side, and then the hypotenuse opposite the 90 degree side. And so in general, if you see any triangle that has those ratios, you say, hey, that's a 30-60-90 triangle. Or if you see a triangle that you know is a 30-60-90 triangle, you could say, hey, I know how to figure out one of the sides based on this ratio right over here. And just as an example, if you see a triangle that looks like this, where the sides are 2, 2 square roots of 3, and 4.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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This right here is the side opposite the 30 degree side, opposite the 60 degree side, and then the hypotenuse opposite the 90 degree side. And so in general, if you see any triangle that has those ratios, you say, hey, that's a 30-60-90 triangle. Or if you see a triangle that you know is a 30-60-90 triangle, you could say, hey, I know how to figure out one of the sides based on this ratio right over here. And just as an example, if you see a triangle that looks like this, where the sides are 2, 2 square roots of 3, and 4. Once again, the ratio of 2 to 2 square roots of 3 is 1 to square root of 3. The ratio of 2 to 4 is the same thing as 1 to 2. This right here must be a 30-60-90 triangle.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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And just as an example, if you see a triangle that looks like this, where the sides are 2, 2 square roots of 3, and 4. Once again, the ratio of 2 to 2 square roots of 3 is 1 to square root of 3. The ratio of 2 to 4 is the same thing as 1 to 2. This right here must be a 30-60-90 triangle. What I want to introduce you to in this video is another important type of triangle that shows up a lot in geometry and a lot in trigonometry, and this is a 45-45-90 triangle. Or another way to think about it is if I have a right triangle that is also isosceles. You obviously can't have a right triangle that is equilateral, because an equilateral triangle has all of their angles have to be 60 degrees, but you can have a right triangle that is isosceles.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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This right here must be a 30-60-90 triangle. What I want to introduce you to in this video is another important type of triangle that shows up a lot in geometry and a lot in trigonometry, and this is a 45-45-90 triangle. Or another way to think about it is if I have a right triangle that is also isosceles. You obviously can't have a right triangle that is equilateral, because an equilateral triangle has all of their angles have to be 60 degrees, but you can have a right triangle that is isosceles. And isosceles, let me write this, this is a right isosceles triangle. And if it's isosceles, that means two of the sides are equal. So these are the two sides that are equal.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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You obviously can't have a right triangle that is equilateral, because an equilateral triangle has all of their angles have to be 60 degrees, but you can have a right triangle that is isosceles. And isosceles, let me write this, this is a right isosceles triangle. And if it's isosceles, that means two of the sides are equal. So these are the two sides that are equal. And then if the two sides are equal, we have proved to ourselves that the base angles are equal. And if we call the measure of these base angles x, then we know that x plus x plus 90 have to be equal to 180. x plus x plus 90 need to be equal to 180. Or if we subtract 90 from both sides, you get x plus x is equal to 90, or 2x is equal to 90.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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So these are the two sides that are equal. And then if the two sides are equal, we have proved to ourselves that the base angles are equal. And if we call the measure of these base angles x, then we know that x plus x plus 90 have to be equal to 180. x plus x plus 90 need to be equal to 180. Or if we subtract 90 from both sides, you get x plus x is equal to 90, or 2x is equal to 90. Or if you divide both sides by 2, you get x is equal to 45 degrees. So a right isosceles triangle can also be called, and this is the more typical name for it, it can also be called a 45-45-90 triangle. And what I want to do in this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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Or if we subtract 90 from both sides, you get x plus x is equal to 90, or 2x is equal to 90. Or if you divide both sides by 2, you get x is equal to 45 degrees. So a right isosceles triangle can also be called, and this is the more typical name for it, it can also be called a 45-45-90 triangle. And what I want to do in this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle. This was actually more straightforward. Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x, and then we can use the Pythagorean theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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And what I want to do in this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle. This was actually more straightforward. Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x, and then we can use the Pythagorean theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. And I'll take the principal root of both sides of that. I want to just change it to yellow, and it keeps not letting me.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. And I'll take the principal root of both sides of that. I want to just change it to yellow, and it keeps not letting me. Okay, to c squared. And I'll take the principal root of both sides of that. The left-hand side, you get principal root of 2 is just square root of 2.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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I want to just change it to yellow, and it keeps not letting me. Okay, to c squared. And I'll take the principal root of both sides of that. The left-hand side, you get principal root of 2 is just square root of 2. And then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2 is equal to c. So if you have a right isosceles triangle, whatever the two legs are, they're going to have the same length. That's why it's isosceles.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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The left-hand side, you get principal root of 2 is just square root of 2. And then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2 is equal to c. So if you have a right isosceles triangle, whatever the two legs are, they're going to have the same length. That's why it's isosceles. The hypotenuse is going to be square root of 2 times that. So c is equal to x times the square root of 2. So, for example, if you have a triangle that looks like this.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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That's why it's isosceles. The hypotenuse is going to be square root of 2 times that. So c is equal to x times the square root of 2. So, for example, if you have a triangle that looks like this. Let me draw it a slightly different way. It's good to have to orient ourselves in different ways every time. So if we see a triangle that's 90 degrees, 45, and 45 like this, and you really just have to know two of these angles to know what the other one is going to be.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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So, for example, if you have a triangle that looks like this. Let me draw it a slightly different way. It's good to have to orient ourselves in different ways every time. So if we see a triangle that's 90 degrees, 45, and 45 like this, and you really just have to know two of these angles to know what the other one is going to be. And if I tell you that this side right over here is 3, I actually don't even have to tell you that this other side is going to be 3. This is an isosceles triangle, so the two legs are going to be the same. And you won't even have to apply the Pythagorean theorem if you know this, and this is a good one to know, that the hypotenuse here, the side opposite the 90-degree side, is just going to be square root of 2 times the length of either of the legs.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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So if we see a triangle that's 90 degrees, 45, and 45 like this, and you really just have to know two of these angles to know what the other one is going to be. And if I tell you that this side right over here is 3, I actually don't even have to tell you that this other side is going to be 3. This is an isosceles triangle, so the two legs are going to be the same. And you won't even have to apply the Pythagorean theorem if you know this, and this is a good one to know, that the hypotenuse here, the side opposite the 90-degree side, is just going to be square root of 2 times the length of either of the legs. So it's going to be 3 times the square root of 2. So the ratio of the sides in a 45-45-90 triangle, or a right isosceles triangle, the ratio of the sides are one of the legs can be 1, then the other leg's going to have the same measure, the same length, and then the hypotenuse is going to be square root of 2 times either of those. 1 to 1, 2 square root of 2.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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And you won't even have to apply the Pythagorean theorem if you know this, and this is a good one to know, that the hypotenuse here, the side opposite the 90-degree side, is just going to be square root of 2 times the length of either of the legs. So it's going to be 3 times the square root of 2. So the ratio of the sides in a 45-45-90 triangle, or a right isosceles triangle, the ratio of the sides are one of the legs can be 1, then the other leg's going to have the same measure, the same length, and then the hypotenuse is going to be square root of 2 times either of those. 1 to 1, 2 square root of 2. So this is 45-45-90. Let me write this as 45-45-90. That's the ratios.
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45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3
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What I'd like to do in this video is use some geometric arguments to prove that the slopes of perpendicular lines are negative reciprocals of each other. And so just to start off, we have lines L and M, and we're going to assume that they are perpendicular, so they intersect at a right angle. We see that depicted right over here. And so I'm gonna now construct some other lines here to help us make our geometric argument. So let me draw a horizontal line that intersects at this point right over here. Let's call that point A. And so let me see if I can do that.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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And so I'm gonna now construct some other lines here to help us make our geometric argument. So let me draw a horizontal line that intersects at this point right over here. Let's call that point A. And so let me see if I can do that. There you go. So that's a horizontal line that intersects at A. And now I'm gonna drop some verticals from that.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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And so let me see if I can do that. There you go. So that's a horizontal line that intersects at A. And now I'm gonna drop some verticals from that. So I'm gonna drop a vertical line right over here, and I'm gonna drop a vertical line right over here. And so that is 90 degrees, and that is 90 degrees, and I've constructed it that way. This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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And now I'm gonna drop some verticals from that. So I'm gonna drop a vertical line right over here, and I'm gonna drop a vertical line right over here. And so that is 90 degrees, and that is 90 degrees, and I've constructed it that way. This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles. And let me now set up some points. So that already said, that's point A. Let's call this point B.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles. And let me now set up some points. So that already said, that's point A. Let's call this point B. Let's call this point C. Let's call this point D. And let's call this point E right over here. Now, let's think about what the slope of line L is. So slope of, let me move this over a little bit.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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Let's call this point B. Let's call this point C. Let's call this point D. And let's call this point E right over here. Now, let's think about what the slope of line L is. So slope of, let me move this over a little bit. So slope, slope of L is going to be what? Well, that's, you could view line L as line, the line that connects point CA. So it's the slope of CA, you could say.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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So slope of, let me move this over a little bit. So slope, slope of L is going to be what? Well, that's, you could view line L as line, the line that connects point CA. So it's the slope of CA, you could say. This is the same thing as slope of line CA. L is line CA. And so to find the slope, that's change in Y over change in X.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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So it's the slope of CA, you could say. This is the same thing as slope of line CA. L is line CA. And so to find the slope, that's change in Y over change in X. So our change in Y is going to be CB. So it's gonna be the length of segment CB. That is our change in Y.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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And so to find the slope, that's change in Y over change in X. So our change in Y is going to be CB. So it's gonna be the length of segment CB. That is our change in Y. So it is CB over our change in X, which is the length of segment BA, which is the length of segment BA right over here. So that is BA. Now, what is the slope of line M?
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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That is our change in Y. So it is CB over our change in X, which is the length of segment BA, which is the length of segment BA right over here. So that is BA. Now, what is the slope of line M? So slope, slope of M, and we could also say slope of, we could call line M line AE, line AE, like that. Well, if we're going to go between point A and point E, once again, it's just change in Y over change in X. Well, what's our change in Y going to be?
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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Now, what is the slope of line M? So slope, slope of M, and we could also say slope of, we could call line M line AE, line AE, like that. Well, if we're going to go between point A and point E, once again, it's just change in Y over change in X. Well, what's our change in Y going to be? Well, our change in Y, well, we're gonna go from this level down to this level as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So we might be tempted to say, well, that's just going to be the length of segment DE.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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Well, what's our change in Y going to be? Well, our change in Y, well, we're gonna go from this level down to this level as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So we might be tempted to say, well, that's just going to be the length of segment DE. But remember, our Y is decreasing. So we're gonna subtract that length as we go from this Y level to that Y level over there. And what is our change in X?
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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So we might be tempted to say, well, that's just going to be the length of segment DE. But remember, our Y is decreasing. So we're gonna subtract that length as we go from this Y level to that Y level over there. And what is our change in X? So our change in X, we're going to go, as we go from A to E, our change in X is going to be the length of segment AD. So AD. So our slope of M is going to be negative DE.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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And what is our change in X? So our change in X, we're going to go, as we go from A to E, our change in X is going to be the length of segment AD. So AD. So our slope of M is going to be negative DE. It's going to be the negative of this length because we're dropping by that much. That's our change in Y over segment A, over segment AD. So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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So our slope of M is going to be negative DE. It's going to be the negative of this length because we're dropping by that much. That's our change in Y over segment A, over segment AD. So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar. And then we're going to be able to show that these are the negative reciprocal of each other. So let's show that these two triangles are similar. So let's say that we have this angle right over here.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar. And then we're going to be able to show that these are the negative reciprocal of each other. So let's show that these two triangles are similar. So let's say that we have this angle right over here. And let's say that angle has measure X, just for kicks. And let's say that we have, let me do another color for, let's say we have this angle right over here. And let's say that the measure, that that has measure Y.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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So let's say that we have this angle right over here. And let's say that angle has measure X, just for kicks. And let's say that we have, let me do another color for, let's say we have this angle right over here. And let's say that the measure, that that has measure Y. Well we know X plus Y plus 90 is equal to 180 because together they are supplementary. So I could write, I could write that X plus 90, plus 90, plus Y, plus Y, is going to be equal to, is going to be equal to 180 degrees. If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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And let's say that the measure, that that has measure Y. Well we know X plus Y plus 90 is equal to 180 because together they are supplementary. So I could write, I could write that X plus 90, plus 90, plus Y, plus Y, is going to be equal to, is going to be equal to 180 degrees. If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees. Is going to be equal to 90 degrees. These are algebraically equivalent statements. So is equal to 90 degrees.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees. Is going to be equal to 90 degrees. These are algebraically equivalent statements. So is equal to 90 degrees. And how can we use this to fill out some of the other angles in these triangles? Well, let's see. X plus this angle down here has to be equal to 90 degrees.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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So is equal to 90 degrees. And how can we use this to fill out some of the other angles in these triangles? Well, let's see. X plus this angle down here has to be equal to 90 degrees. Or you could say X plus 90 plus what is going to be equal to 180. I'm looking at triangle CBA right over here. The interior angles of a triangle add up to 180.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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X plus this angle down here has to be equal to 90 degrees. Or you could say X plus 90 plus what is going to be equal to 180. I'm looking at triangle CBA right over here. The interior angles of a triangle add up to 180. So X plus 90 plus what is equal to 180? Well X plus 90 plus Y is equal to 180. We already established that.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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The interior angles of a triangle add up to 180. So X plus 90 plus what is equal to 180? Well X plus 90 plus Y is equal to 180. We already established that. Similarly over here. Y plus 90 plus what is going to be equal to 180? Well, same argument.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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We already established that. Similarly over here. Y plus 90 plus what is going to be equal to 180? Well, same argument. We already know Y plus 90 plus X is equal to 180. So Y plus 90 plus X is equal to, is equal to 180. And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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Well, same argument. We already know Y plus 90 plus X is equal to 180. So Y plus 90 plus X is equal to, is equal to 180. And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other. They both have an angle of X, they both have a measure X, they both have an angle of measure Y, and they're both right triangles. So just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates, we know that triangle E, triangle EDA, EDA is similar to triangle, to triangle ABC, to triangle ABC. And so that tells us that the ratio of corresponding sides are going to be the same.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other. They both have an angle of X, they both have a measure X, they both have an angle of measure Y, and they're both right triangles. So just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates, we know that triangle E, triangle EDA, EDA is similar to triangle, to triangle ABC, to triangle ABC. And so that tells us that the ratio of corresponding sides are going to be the same. And so for example, we know, let's find the ratio of corresponding sides. We know that the ratio of, let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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And so that tells us that the ratio of corresponding sides are going to be the same. And so for example, we know, let's find the ratio of corresponding sides. We know that the ratio of, let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So the ratio of CB over BA, over BA is going to be equal to, is going to be equal to, well the corresponding side to CB, it's the side opposite the X degree angle right over here. So the corresponding side to CB is side AD, so that's going to be equal to AD over, what's the corresponding side to BA? Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So the ratio of CB over BA, over BA is going to be equal to, is going to be equal to, well the corresponding side to CB, it's the side opposite the X degree angle right over here. So the corresponding side to CB is side AD, so that's going to be equal to AD over, what's the corresponding side to BA? Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE. And so this right over here, this right over here we saw from the beginning, this is the slope, this is the slope of L. So slope, slope of L. And how does this relate to the slope of M? Notice the slope of M is the negative reciprocal of this. You take the reciprocal, you're going to get DE over AD, and then you have to take this negative right over here.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE. And so this right over here, this right over here we saw from the beginning, this is the slope, this is the slope of L. So slope, slope of L. And how does this relate to the slope of M? Notice the slope of M is the negative reciprocal of this. You take the reciprocal, you're going to get DE over AD, and then you have to take this negative right over here. So we could write this as the negative reciprocal of slope of M. Negative reciprocal, reciprocal of M's, of M's slope. And there you have it. We've just shown that if we start with, if we assume these L and M are perpendicular, and we set up these similar triangles, and we were able to show that the slope of L is the negative reciprocal of the slope of M.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3
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We see that we're leaving out less We see when we have now. This is a one two three four five six seven sided polygon We're leaving we're leaving a little bit less We're underestimating still but we're underestimating by less this area that we're giving up isn't as large as this area right over here So in this approximation we have What I say seven triangles one two three four five six seven triangles and the area of each of those triangles is once again a b over two now a and b here are different than a and b over here and notice what's happening as As we increase as we increase the number of triangles Not only is it approximating the area of the circle better But a is getting longer And you can see as you could imagine is as we increase many many many more triangles a is going to approach our Now another thing to think about is what is seven? What is seven times B approaching? so we're saying that a is approaching R as we add more of these sides of the polygon as we add more triangles and what is the number of Triangles times the base of the triangle. What is that approaching? Well, this is going to approach is going to approach the perimeter of the pot Or this is going to be the perimeter of the polygon. So seven times B is that Plus let me actually let me draw this it's that plus that Plus that you get the point plus that plus that plus that plus that So once again Seven let me write this as seven times B.
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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so we're saying that a is approaching R as we add more of these sides of the polygon as we add more triangles and what is the number of Triangles times the base of the triangle. What is that approaching? Well, this is going to approach is going to approach the perimeter of the pot Or this is going to be the perimeter of the polygon. So seven times B is that Plus let me actually let me draw this it's that plus that Plus that you get the point plus that plus that plus that plus that So once again Seven let me write this as seven times B. That is the perimeter of the polygon perimeter Perimeter of the polygon So think about what's happening as we as we have more and more sides of the polygon our a Our height of each of our triangles is going to approach Our radius is going to approach The radius is going to get longed the height of each of the triangles gonna get longer and longer It's going to approach the radius as we have as we approach an infinite number of triangles And then the number of polygons we have are again the number of polygons the number of sides we have Times the bases that's going to be the perimeter of the polygon and as we add more and more sides as we add more and the perimeter of the polygon is going to approach the circumference of the circle. I'll write it out, circumference. And you see that even more clearly right over here.
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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So seven times B is that Plus let me actually let me draw this it's that plus that Plus that you get the point plus that plus that plus that plus that So once again Seven let me write this as seven times B. That is the perimeter of the polygon perimeter Perimeter of the polygon So think about what's happening as we as we have more and more sides of the polygon our a Our height of each of our triangles is going to approach Our radius is going to approach The radius is going to get longed the height of each of the triangles gonna get longer and longer It's going to approach the radius as we have as we approach an infinite number of triangles And then the number of polygons we have are again the number of polygons the number of sides we have Times the bases that's going to be the perimeter of the polygon and as we add more and more sides as we add more and the perimeter of the polygon is going to approach the circumference of the circle. I'll write it out, circumference. And you see that even more clearly right over here. So once again, how many sides do I have here? I have one, two, three, four, five, six, seven, eight, nine, ten sides. So this, I can write the perimeter of the polygon as ten times b, and then if I multiply that times a over two, if I multiply that times this another color, a over, let me just write it like this, times a over two, I'm once again approximating the area of the circle, because a times b over two, that's the area of each of these triangles, and then I have ten of these triangles.
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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And you see that even more clearly right over here. So once again, how many sides do I have here? I have one, two, three, four, five, six, seven, eight, nine, ten sides. So this, I can write the perimeter of the polygon as ten times b, and then if I multiply that times a over two, if I multiply that times this another color, a over, let me just write it like this, times a over two, I'm once again approximating the area of the circle, because a times b over two, that's the area of each of these triangles, and then I have ten of these triangles. But now let's think about this more generally. Let's think about it if I were to have n, if I were to have an n-sided polygon. So I have an n-sided polygon, then I'd be approximating the area as n times b, n times b.
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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So this, I can write the perimeter of the polygon as ten times b, and then if I multiply that times a over two, if I multiply that times this another color, a over, let me just write it like this, times a over two, I'm once again approximating the area of the circle, because a times b over two, that's the area of each of these triangles, and then I have ten of these triangles. But now let's think about this more generally. Let's think about it if I were to have n, if I were to have an n-sided polygon. So I have an n-sided polygon, then I'd be approximating the area as n times b, n times b. We see this right over here, when n is equal to ten, we have ten times b, so it's n times b times a over two. This isn't something mysterious. The base times the height divided by two, this right over here, that's the area of each triangle, and then I'm going to have n of these triangles.
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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So I have an n-sided polygon, then I'd be approximating the area as n times b, n times b. We see this right over here, when n is equal to ten, we have ten times b, so it's n times b times a over two. This isn't something mysterious. The base times the height divided by two, this right over here, that's the area of each triangle, and then I'm going to have n of these triangles. So this is our approximation for the area, so let me write this, the area is going to be approximately that right over there. It's going to be n, the number of triangles I have, times the area of each triangle. Now what's going to happen as n approaches infinity, as I approach having an infinite-sided polygon, as I have an infinite number of triangles?
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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The base times the height divided by two, this right over here, that's the area of each triangle, and then I'm going to have n of these triangles. So this is our approximation for the area, so let me write this, the area is going to be approximately that right over there. It's going to be n, the number of triangles I have, times the area of each triangle. Now what's going to happen as n approaches infinity, as I approach having an infinite-sided polygon, as I have an infinite number of triangles? So let's just think this through a little bit, because this is where it gets interesting. This is the informal argument. To do this better, I'd have to dig out a little bit of calculus, but this gives you the essence.
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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Now what's going to happen as n approaches infinity, as I approach having an infinite-sided polygon, as I have an infinite number of triangles? So let's just think this through a little bit, because this is where it gets interesting. This is the informal argument. To do this better, I'd have to dig out a little bit of calculus, but this gives you the essence. So let's just think about what happens as n approaches infinity. So as n approaches infinity, we've already said, as we have more and more sides, and we have more and more triangles, more and more triangles, a approaches r. So let's write that down. So a is going to approach r, the height of the triangles is going to approach the radius.
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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To do this better, I'd have to dig out a little bit of calculus, but this gives you the essence. So let's just think about what happens as n approaches infinity. So as n approaches infinity, we've already said, as we have more and more sides, and we have more and more triangles, more and more triangles, a approaches r. So let's write that down. So a is going to approach r, the height of the triangles is going to approach the radius. And what else is going to happen? Well, n times b, the perimeter of the polygon, the perimeter of the polygon, is going to approach the circumference. So a is going to approach r, and n times b is going to approach the circumference.
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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So a is going to approach r, the height of the triangles is going to approach the radius. And what else is going to happen? Well, n times b, the perimeter of the polygon, the perimeter of the polygon, is going to approach the circumference. So a is going to approach r, and n times b is going to approach the circumference. Or another way of thinking about it, if it's approaching the circumference, we could say that n times b is going to approach 2 pi times the radius, because that's what the circumference is going to be equal to. So if a is approaching the radius, and nb is approaching 2 pi r, well then, what is the area of our polygon? Or what is the area of our polygon going to, or the area of our circle going to be?
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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So a is going to approach r, and n times b is going to approach the circumference. Or another way of thinking about it, if it's approaching the circumference, we could say that n times b is going to approach 2 pi times the radius, because that's what the circumference is going to be equal to. So if a is approaching the radius, and nb is approaching 2 pi r, well then, what is the area of our polygon? Or what is the area of our polygon going to, or the area of our circle going to be? Well, it is going to approach, or I should say the area of our polygon is going to approach, nb is going to approach 2 pi r, so instead of nb, I'm writing 2 pi r there. a is going to approach r, and then I'm dividing it by 2. So as n approaches infinity, as we have an infinite number of sides of our polygon, an infinite number of triangles, the area of our polygon will approach this, which is equal to what?
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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Or what is the area of our polygon going to, or the area of our circle going to be? Well, it is going to approach, or I should say the area of our polygon is going to approach, nb is going to approach 2 pi r, so instead of nb, I'm writing 2 pi r there. a is going to approach r, and then I'm dividing it by 2. So as n approaches infinity, as we have an infinite number of sides of our polygon, an infinite number of triangles, the area of our polygon will approach this, which is equal to what? Well, you have 2 divided by 2, and then pi r times r is equal to pi r squared. So as we approach having an infinite number of triangles, an infinite number of sides, we see that we approach the area of the circle. And as we approach the area of the circle, we are approaching pi r squared.
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Area of a circle intuition High School Geometry High School Math Khan Academy.mp3
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We're told that Shui concluded the quadrilaterals, these two over here, have four pairs of congruent corresponding angles. We can see these right over there. And so based on that, she concludes that the figures are similar. What error, if any, did Shui make in her conclusion? Pause this video and try to figure this out on your own. All right, so let's just remind ourselves one definition of similarity that we often use in geometry class, and that's two figures are similar as if you can, through a series of rigid transformations and dilations, if you can map one figure on to the other. Now, when I look at these two figures, you could try to do something.
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Similar shapes & transformations.mp3
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What error, if any, did Shui make in her conclusion? Pause this video and try to figure this out on your own. All right, so let's just remind ourselves one definition of similarity that we often use in geometry class, and that's two figures are similar as if you can, through a series of rigid transformations and dilations, if you can map one figure on to the other. Now, when I look at these two figures, you could try to do something. You could say, okay, let me shift it so that k gets mapped on to h, and if you did that, it looks like l would get mapped on to g, but these sides, kn and lm right over here, they seem a good bit longer. So, and then if you tried to dilate it down so that the length of kn is the same as the length of hi, well then, the lengths of kl and gh would be different. So it doesn't seem like you can do this.
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Similar shapes & transformations.mp3
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Now, when I look at these two figures, you could try to do something. You could say, okay, let me shift it so that k gets mapped on to h, and if you did that, it looks like l would get mapped on to g, but these sides, kn and lm right over here, they seem a good bit longer. So, and then if you tried to dilate it down so that the length of kn is the same as the length of hi, well then, the lengths of kl and gh would be different. So it doesn't seem like you can do this. So it is strange that Shui concluded that they are similar. So let's find the mistake. I'm already, I'll already rule out c, that it's a correct conclusion, because I don't think they are similar.
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Similar shapes & transformations.mp3
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So it doesn't seem like you can do this. So it is strange that Shui concluded that they are similar. So let's find the mistake. I'm already, I'll already rule out c, that it's a correct conclusion, because I don't think they are similar. So let's see. Is the error that a rigid transformation, a translation, would map hg on to kl? Yep, we just talked about that.
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Similar shapes & transformations.mp3
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I'm already, I'll already rule out c, that it's a correct conclusion, because I don't think they are similar. So let's see. Is the error that a rigid transformation, a translation, would map hg on to kl? Yep, we just talked about that. Hg can be mapped on to kl. So the quadrilaterals are congruent, not similar. Oh, choice A is making an even stronger statement, because anything that is congruent is going to be similar.
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Similar shapes & transformations.mp3
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Yep, we just talked about that. Hg can be mapped on to kl. So the quadrilaterals are congruent, not similar. Oh, choice A is making an even stronger statement, because anything that is congruent is going to be similar. You actually can't have something that's congruent and not similar, and so choice A does not make any sense. So our deductive reasoning tells us it's probably choice B, but let's just read it. It's impossible to map quadrilateral ghij onto quadrilateral lknm using only rigid transformations and dilations, so the figures are not similar.
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Similar shapes & transformations.mp3
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Oh, choice A is making an even stronger statement, because anything that is congruent is going to be similar. You actually can't have something that's congruent and not similar, and so choice A does not make any sense. So our deductive reasoning tells us it's probably choice B, but let's just read it. It's impossible to map quadrilateral ghij onto quadrilateral lknm using only rigid transformations and dilations, so the figures are not similar. Yeah, that's right. You could try. You could map hg on to kl, but then segment ij would look something like this.
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Similar shapes & transformations.mp3
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It's impossible to map quadrilateral ghij onto quadrilateral lknm using only rigid transformations and dilations, so the figures are not similar. Yeah, that's right. You could try. You could map hg on to kl, but then segment ij would look something like this. Ij would go right over here. And then if you tried to dilate it so that the length of hi and gj matched kn or lm, then you're going to make hg bigger as well, so you're never going to be able to map them on to each other, even if you can use dilations. So I like choice B.
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Similar shapes & transformations.mp3
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So pause this video and see if you can figure this out on your own. All right, now before I even look at the choices, I like to just think about what would that dilation actually look like? So our center of dilation is P and it's a scale factor of 3 4th. So one way to think about it is, however far any point was from P before, it's not going to be 3 4ths as far, but along the same line. So I'm just going to estimate it. So if C was there, 3 1 2 would be this far. So 3 4ths would be right about there.
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Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3
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So one way to think about it is, however far any point was from P before, it's not going to be 3 4ths as far, but along the same line. So I'm just going to estimate it. So if C was there, 3 1 2 would be this far. So 3 4ths would be right about there. So C prime should be about there. If we have this line connecting B and P like this, 3, let's see, half of that is there. 3 4ths is going to be there.
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Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3
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So 3 4ths would be right about there. So C prime should be about there. If we have this line connecting B and P like this, 3, let's see, half of that is there. 3 4ths is going to be there. So B prime should be there. And then on this line, halfway is roughly there. I'm just eyeballing it.
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Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3
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3 4ths is going to be there. So B prime should be there. And then on this line, halfway is roughly there. I'm just eyeballing it. So 3 4ths is there. So A prime, A prime should be there. And so A prime, B prime, C prime should look something like this, which we can see is exactly what we see for choice C. So choice C, it looks correct.
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Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3
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I'm just eyeballing it. So 3 4ths is there. So A prime, A prime should be there. And so A prime, B prime, C prime should look something like this, which we can see is exactly what we see for choice C. So choice C, it looks correct. So I'm gonna just circle that or select it just like that. But let's just make sure we understand why these other two choices were not correct. So choice A, it looks like it is a dilation with a 3 4ths scale factor.
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Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3
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And so A prime, B prime, C prime should look something like this, which we can see is exactly what we see for choice C. So choice C, it looks correct. So I'm gonna just circle that or select it just like that. But let's just make sure we understand why these other two choices were not correct. So choice A, it looks like it is a dilation with a 3 4ths scale factor. Each of the dimensions, each of the sides of these triangles, of this triangle looks like it's about 3 4ths of what it originally was. But it doesn't look like the center of dilation is P. Here, the center of dilation looks like it is probably the midpoint between or the midpoint of segment AC, because now it looks like everything is 3 4ths of the distance it was to that point. So they have this other center of dilation in choice A.
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Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3
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So choice A, it looks like it is a dilation with a 3 4ths scale factor. Each of the dimensions, each of the sides of these triangles, of this triangle looks like it's about 3 4ths of what it originally was. But it doesn't look like the center of dilation is P. Here, the center of dilation looks like it is probably the midpoint between or the midpoint of segment AC, because now it looks like everything is 3 4ths of the distance it was to that point. So they have this other center of dilation in choice A. The center of dilation is not P, and that's why we can rule that one out. And then for choice B right over here, it looks like they just got the scale factor wrong. Actually, they got the center of dilation and the scale factor wrong.
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Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3
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So they have this other center of dilation in choice A. The center of dilation is not P, and that's why we can rule that one out. And then for choice B right over here, it looks like they just got the scale factor wrong. Actually, they got the center of dilation and the scale factor wrong. It still looks like they are using this as a center of dilation, but this scale factor looks like it's much closer to 1 4th or 1 3rd, not 3 4ths. So that's why we can rule that one out as well. We like our choice C.
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Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3
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Well, let's say we wanted to construct a square. How could we slice a cube with a plane to get the intersection of this cube and that plane to be a square? Well, imagine if that plane were to cut just like this. A square is maybe the most obvious one. So it cuts the top right over there. It cuts the top right over there. It cuts this side right over here.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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A square is maybe the most obvious one. So it cuts the top right over there. It cuts the top right over there. It cuts this side right over here. It cuts this side right over here. It cuts this side, I guess, in the back. If it's a glass cube, you'd be able to see it.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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It cuts this side right over here. It cuts this side right over here. It cuts this side, I guess, in the back. If it's a glass cube, you'd be able to see it. Right over there, a dotted line. And then it cuts this right over here. So you could imagine a plane that did this.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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If it's a glass cube, you'd be able to see it. Right over there, a dotted line. And then it cuts this right over here. So you could imagine a plane that did this. And if I wanted to draw the broader plane, I could draw it like this. Let me see if I can do a decent and adequate job of drawing the actual, I guess you could say, part of the plane that is cutting this cube. It could look something like this.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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So you could imagine a plane that did this. And if I wanted to draw the broader plane, I could draw it like this. Let me see if I can do a decent and adequate job of drawing the actual, I guess you could say, part of the plane that is cutting this cube. It could look something like this. And I could even color in the part of the plane that you could actually see if the cube were opaque. If you couldn't see through it. But if you could see it through it, you'd see this dotted line and the plane would look like that.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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It could look something like this. And I could even color in the part of the plane that you could actually see if the cube were opaque. If you couldn't see through it. But if you could see it through it, you'd see this dotted line and the plane would look like that. So a square is a pretty straightforward thing to get if you're doing a planar slice of a cube. But what about a rectangle? How can you get that?
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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But if you could see it through it, you'd see this dotted line and the plane would look like that. So a square is a pretty straightforward thing to get if you're doing a planar slice of a cube. But what about a rectangle? How can you get that? And at any point, I encourage you, pause the video and try to think about it on your own. How can you get these shapes that I'm talking about? Well, a rectangle, you can actually cut like this.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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How can you get that? And at any point, I encourage you, pause the video and try to think about it on your own. How can you get these shapes that I'm talking about? Well, a rectangle, you can actually cut like this. So if you cut this side like this, and then cut that side like that, and then you cut this side like that, I think you see where this is going, this side like that, and then you cut the bottom right over there, then the intersection of the plane that you're cutting with, so the intersection, let's see, this could be the plane that I'm actually cutting with, so the intersection of the plane that I'm cutting with and my cube is going to be a rectangle. So it might look like this, and once again, let me shade in the stuff. If you kind of view this, if you imagine the plane is like one of those huge blades that magicians use to saw people in half, or pretend like they're, or give us the illusion of sawing people in half, it might look something like this.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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Well, a rectangle, you can actually cut like this. So if you cut this side like this, and then cut that side like that, and then you cut this side like that, I think you see where this is going, this side like that, and then you cut the bottom right over there, then the intersection of the plane that you're cutting with, so the intersection, let's see, this could be the plane that I'm actually cutting with, so the intersection of the plane that I'm cutting with and my cube is going to be a rectangle. So it might look like this, and once again, let me shade in the stuff. If you kind of view this, if you imagine the plane is like one of those huge blades that magicians use to saw people in half, or pretend like they're, or give us the illusion of sawing people in half, it might look something like this. Okay, so you're like, okay, that's not so hard to digest, that I can, if I intersect a plane with a cube, I can get a square, I can get a rectangle. But what about triangles? Well, once again, and pause the video if you think you can figure it out, triangles, not so bad.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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If you kind of view this, if you imagine the plane is like one of those huge blades that magicians use to saw people in half, or pretend like they're, or give us the illusion of sawing people in half, it might look something like this. Okay, so you're like, okay, that's not so hard to digest, that I can, if I intersect a plane with a cube, I can get a square, I can get a rectangle. But what about triangles? Well, once again, and pause the video if you think you can figure it out, triangles, not so bad. You could cut this side right over here, this side right over here, and this side right over here, and then this is, and of course I could keep drawing the plane, but I think you get the idea. This would be a triangle, and there's different types of triangles that you can construct. You could construct an equilateral triangle, so as long as this cut is the same length as this cut right over here, is the same length as, or the length that it intersects on this space of the cube, that's going to be an equilateral triangle.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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Well, once again, and pause the video if you think you can figure it out, triangles, not so bad. You could cut this side right over here, this side right over here, and this side right over here, and then this is, and of course I could keep drawing the plane, but I think you get the idea. This would be a triangle, and there's different types of triangles that you can construct. You could construct an equilateral triangle, so as long as this cut is the same length as this cut right over here, is the same length as, or the length that it intersects on this space of the cube, that's going to be an equilateral triangle. If you pushed this point out more, actually let me do that in a different color, if you pushed it out more, you're going to have an isosceles triangle. You're going to have an isosceles triangle. If you were to bring this point really, really, really close, like here, you would approach having a right angle, but it wouldn't be quite a right angle.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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You could construct an equilateral triangle, so as long as this cut is the same length as this cut right over here, is the same length as, or the length that it intersects on this space of the cube, that's going to be an equilateral triangle. If you pushed this point out more, actually let me do that in a different color, if you pushed it out more, you're going to have an isosceles triangle. You're going to have an isosceles triangle. If you were to bring this point really, really, really close, like here, you would approach having a right angle, but it wouldn't be quite a right angle. You'd still have, these angles would still be less than 90 degrees. You can approach 90 degrees, so you can't quite have an exactly a right angle. And so you can't get to 90 degrees, you're definitely not going to get to 91 degrees, so you're actually not going to be able to do an obtuse triangle either.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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If you were to bring this point really, really, really close, like here, you would approach having a right angle, but it wouldn't be quite a right angle. You'd still have, these angles would still be less than 90 degrees. You can approach 90 degrees, so you can't quite have an exactly a right angle. And so you can't get to 90 degrees, you're definitely not going to get to 91 degrees, so you're actually not going to be able to do an obtuse triangle either. But you can do an equilateral, you can do an isosceles, you can do scalene triangles. You could do the different types of acute triangles. But now let's do some really interesting things.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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And so you can't get to 90 degrees, you're definitely not going to get to 91 degrees, so you're actually not going to be able to do an obtuse triangle either. But you can do an equilateral, you can do an isosceles, you can do scalene triangles. You could do the different types of acute triangles. But now let's do some really interesting things. Can you get a pentagon by slicing a cube with a plane? And I really want you to pause the video and think about it here, because that's just a fun thing to think about. How can you get a pentagon by slicing a cube with a plane?
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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But now let's do some really interesting things. Can you get a pentagon by slicing a cube with a plane? And I really want you to pause the video and think about it here, because that's just a fun thing to think about. How can you get a pentagon by slicing a cube with a plane? All right, so here I go. This is how you can get a pentagon by slicing a cube with a plane. Imagine slicing the top, let me do it a little bit different.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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How can you get a pentagon by slicing a cube with a plane? All right, so here I go. This is how you can get a pentagon by slicing a cube with a plane. Imagine slicing the top, let me do it a little bit different. So imagine slicing the top right over there. So we imagine slicing the top like this. But imagine slicing this back side like that.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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Imagine slicing the top, let me do it a little bit different. So imagine slicing the top right over there. So we imagine slicing the top like this. But imagine slicing this back side like that. This back side that you can't see quite like that. Now you slice this side right over here like this. And then you slice this side right over here like this.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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But imagine slicing this back side like that. This back side that you can't see quite like that. Now you slice this side right over here like this. And then you slice this side right over here like this. This could be, if I wanted to draw the plane, it maybe won't be so obvious if I try to draw the plane, but you get the actual idea. If I slice the right angle, not at a right angle, at the right, at a right angle, actually I shouldn't use the word right angle, that'll confuse everything. If I slice it in the proper angle, then the intersection of my plane and my cube is going to be this pentagon right over here.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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And then you slice this side right over here like this. This could be, if I wanted to draw the plane, it maybe won't be so obvious if I try to draw the plane, but you get the actual idea. If I slice the right angle, not at a right angle, at the right, at a right angle, actually I shouldn't use the word right angle, that'll confuse everything. If I slice it in the proper angle, then the intersection of my plane and my cube is going to be this pentagon right over here. Now let's up the stakes something. Let's up the stakes even more. What about a hexagon?
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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If I slice it in the proper angle, then the intersection of my plane and my cube is going to be this pentagon right over here. Now let's up the stakes something. Let's up the stakes even more. What about a hexagon? Can I slice a cube in a way with a two-dimensional plane to get the intersection of the plane and the cube being a hexagon? Well, as you can imagine, I wouldn't have asked you the question unless I could. So let's see if we can do it.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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What about a hexagon? Can I slice a cube in a way with a two-dimensional plane to get the intersection of the plane and the cube being a hexagon? Well, as you can imagine, I wouldn't have asked you the question unless I could. So let's see if we can do it. If we slice this right over there, if we slice this bottom piece right over there, and then you slice this back side like that, you slice this back side like that, and then you slice this side that we can see right over there, and then this side that we can see right over there, and this side I could have written it much straighter. Hopefully you get the idea. I can slice this cube so that I can actually get a hexagon.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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So let's see if we can do it. If we slice this right over there, if we slice this bottom piece right over there, and then you slice this back side like that, you slice this back side like that, and then you slice this side that we can see right over there, and then this side that we can see right over there, and this side I could have written it much straighter. Hopefully you get the idea. I can slice this cube so that I can actually get a hexagon. Hopefully this gives you a better appreciation for what you can actually do with a cube, especially if you're busy slicing it with large planar planes or large planar blades in some way. There's actually more to a cube than maybe you might have imagined in the past. One way to think about it is there are six sides to a cube or six surfaces to a cube.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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I can slice this cube so that I can actually get a hexagon. Hopefully this gives you a better appreciation for what you can actually do with a cube, especially if you're busy slicing it with large planar planes or large planar blades in some way. There's actually more to a cube than maybe you might have imagined in the past. One way to think about it is there are six sides to a cube or six surfaces to a cube. You can cut as many as six of the surfaces when you intersect it with a plane, and every time you cut into one of those surfaces, it forms a side. Here we're cutting into four sides. Here we're cutting into four surfaces or four sides.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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One way to think about it is there are six sides to a cube or six surfaces to a cube. You can cut as many as six of the surfaces when you intersect it with a plane, and every time you cut into one of those surfaces, it forms a side. Here we're cutting into four sides. Here we're cutting into four surfaces or four sides. Here we're cutting into three. Here we're cutting into five. We're not cutting into the bottom of the cube here.
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Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3
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And then they say use one of the triangles, so use one of these three triangles, to approximate the ratio, the ratio is the length of segment PN divided by the length of segment MN. So they want us to figure out the ratio PN over MN. So pause this video and see if you can figure this out. All right, now let's work through this together. Now, given that they want us to figure out this ratio, and they want us to actually evaluate it or be able to approximate it, we are probably dealing with similarity. And so what I would wanna look for is, are one of these triangles similar to the triangle we have here? And we're dealing with similar triangles if we have two angles in common, because if we have two angles in common, then that means that we definitely have the third angle as well, because the third angle is completely determined by what the other two angles are.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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All right, now let's work through this together. Now, given that they want us to figure out this ratio, and they want us to actually evaluate it or be able to approximate it, we are probably dealing with similarity. And so what I would wanna look for is, are one of these triangles similar to the triangle we have here? And we're dealing with similar triangles if we have two angles in common, because if we have two angles in common, then that means that we definitely have the third angle as well, because the third angle is completely determined by what the other two angles are. So we have a 35 degree angle here, and we have a 90 degree angle here. And of all of these choices, this doesn't have a 35 degree angle, it has a 90. This doesn't have a 35, it has a 90.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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And we're dealing with similar triangles if we have two angles in common, because if we have two angles in common, then that means that we definitely have the third angle as well, because the third angle is completely determined by what the other two angles are. So we have a 35 degree angle here, and we have a 90 degree angle here. And of all of these choices, this doesn't have a 35 degree angle, it has a 90. This doesn't have a 35, it has a 90. But triangle two here has a 35 degree angle, has a 90 degree angle, and has a 55 degree angle. And if you did the math, knowing that 35 plus 90 plus this have to add up to 180 degrees, you would see that this too has a measure of 55 degrees. And so given that all of our angle measures are the same between triangle PNM and triangle number two right over here, we know that these two are similar triangles.
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Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3
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