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So there's a bunch of things we know about vertical angles and angles of transversals. The most obvious one is that we have this vertical. We know that angle AEB is going to be congruent, or its measure is going to be equal to the measure of angle CED. So we know that angle AEB is going to be congruent to angle DEC, which really just means they have the exact same measure. And we know that because they are vertical angles. Now we also know that AB and CD are parallel, so this line right over here is a transversal. So we know, for example, and there's actually several ways that we can do this problem, but we know that this is a transversal, and there's a couple of ways to think about it right over here.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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So we know that angle AEB is going to be congruent to angle DEC, which really just means they have the exact same measure. And we know that because they are vertical angles. Now we also know that AB and CD are parallel, so this line right over here is a transversal. So we know, for example, and there's actually several ways that we can do this problem, but we know that this is a transversal, and there's a couple of ways to think about it right over here. So let me just continue the transversal so we get to see all of the different angles. You could say that this angle right here, angle ABE, so this is its measure right over here, you could say that it is the alternate interior angle to angle ECD, to this angle right over there. And if that didn't jump out of view, you would say that the corresponding angle to this one right over here is this angle right up here.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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So we know, for example, and there's actually several ways that we can do this problem, but we know that this is a transversal, and there's a couple of ways to think about it right over here. So let me just continue the transversal so we get to see all of the different angles. You could say that this angle right here, angle ABE, so this is its measure right over here, you could say that it is the alternate interior angle to angle ECD, to this angle right over there. And if that didn't jump out of view, you would say that the corresponding angle to this one right over here is this angle right up here. If you were to continue this line off a little bit, these are the corresponding angles, and then this one is vertical. But either way, angle ABE is going to be congruent to angle DCE. And we could say because it's alternate interior angles.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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And if that didn't jump out of view, you would say that the corresponding angle to this one right over here is this angle right up here. If you were to continue this line off a little bit, these are the corresponding angles, and then this one is vertical. But either way, angle ABE is going to be congruent to angle DCE. And we could say because it's alternate interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle, and then the next side is congruent to the next side over here. So pink-green side.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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And we could say because it's alternate interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle, and then the next side is congruent to the next side over here. So pink-green side. So we can employ AAS, angle-angle side. And it's in the right order. So now we know that triangle, we have to make sure that we get the letters right here, that we have the right corresponding vertices.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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So pink-green side. So we can employ AAS, angle-angle side. And it's in the right order. So now we know that triangle, we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB, triangle AE, actually let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle going to the green angle and then going to the one that we haven't labeled. So angle BEA we can say is congruent to angle, we start with the magenta vertices, C, go to the center, E, and then go to the unlabeled one, D. And we know this because of angle-angle side and they correspond to each other.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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So now we know that triangle, we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB, triangle AE, actually let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle going to the green angle and then going to the one that we haven't labeled. So angle BEA we can say is congruent to angle, we start with the magenta vertices, C, go to the center, E, and then go to the unlabeled one, D. And we know this because of angle-angle side and they correspond to each other. Magenta-green side, magenta-green side, they're all congruent. So this is from AAS. And then if we know that they are congruent, that means corresponding sides are congruent.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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So angle BEA we can say is congruent to angle, we start with the magenta vertices, C, go to the center, E, and then go to the unlabeled one, D. And we know this because of angle-angle side and they correspond to each other. Magenta-green side, magenta-green side, they're all congruent. So this is from AAS. And then if we know that they are congruent, that means corresponding sides are congruent. So then we know that this side, so we know these two triangles are congruent, so that means that the corresponding sides are congruent. So then we know that length of BE, that we know that BE, the length of that segment BE is going to be equal. And that's the segment that's between the magenta and the green angles.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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And then if we know that they are congruent, that means corresponding sides are congruent. So then we know that this side, so we know these two triangles are congruent, so that means that the corresponding sides are congruent. So then we know that length of BE, that we know that BE, the length of that segment BE is going to be equal. And that's the segment that's between the magenta and the green angles. The corresponding side is side CE, between the magenta and the green angles, is equal to CE. And this just comes out of the previous statement. If we number them, that's 1, that's 2, and that's 3.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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And that's the segment that's between the magenta and the green angles. The corresponding side is side CE, between the magenta and the green angles, is equal to CE. And this just comes out of the previous statement. If we number them, that's 1, that's 2, and that's 3. And so that comes out of statement 3. And so we have proven this. E is the midpoint of BC.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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If we number them, that's 1, that's 2, and that's 3. And so that comes out of statement 3. And so we have proven this. E is the midpoint of BC. It comes straight out of the fact that BE is equal to CE. So I can mark this off with a hash. This line segment right over here is congruent to this line segment right over here because we know that those two triangles are congruent.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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E is the midpoint of BC. It comes straight out of the fact that BE is equal to CE. So I can mark this off with a hash. This line segment right over here is congruent to this line segment right over here because we know that those two triangles are congruent. And I've inadvertently right here done a little two-column proof. This over here on the left-hand side is my statement. And then on the right-hand side, I gave my reason.
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Congruent triangle proof example Congruence Geometry Khan Academy.mp3
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It's something that would store grain and then it can kind of fall out of the bottom, has a radius of 10 meters at the top and is eight meters tall. So let's draw that. So it's cone-shaped and it has a radius at the top. So the top must be where the base is. I guess one way to think about it, it must be the wider part of the cone. So it looks like this, something like that. That's what this first sentence tells us.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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So the top must be where the base is. I guess one way to think about it, it must be the wider part of the cone. So it looks like this, something like that. That's what this first sentence tells us. It has a radius of 10 meters. So this distance right over here is 10 meters. And the height is eight meters.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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That's what this first sentence tells us. It has a radius of 10 meters. So this distance right over here is 10 meters. And the height is eight meters. They say it's eight meters tall. So this right over here is eight meters. Then they tell us it is filled up to two meters from the top with grain.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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And the height is eight meters. They say it's eight meters tall. So this right over here is eight meters. Then they tell us it is filled up to two meters from the top with grain. So one way to think about it, it's filled about this high with grain. So it's filled about that high with grain. So this distance is going to be eight minus these two.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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Then they tell us it is filled up to two meters from the top with grain. So one way to think about it, it's filled about this high with grain. So it's filled about that high with grain. So this distance is going to be eight minus these two. So this is going to be six meters high. That's what that second sentence tells us. The hopper will pour the grain into boxes with dimensions of 0.5 meters by 0.5 meters by 0.4 meters.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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So this distance is going to be eight minus these two. So this is going to be six meters high. That's what that second sentence tells us. The hopper will pour the grain into boxes with dimensions of 0.5 meters by 0.5 meters by 0.4 meters. The hopper pours grain at a rate of eight cubic meters per minute. So the first, a lot of information there. The first question is what is the volume of grain in the hopper?
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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The hopper will pour the grain into boxes with dimensions of 0.5 meters by 0.5 meters by 0.4 meters. The hopper pours grain at a rate of eight cubic meters per minute. So the first, a lot of information there. The first question is what is the volume of grain in the hopper? So before we even get to these other questions, let's see if we can answer that. So that's going to be this volume right over here of the red part, the cone made up of the grain. Pause this video and try to figure it out.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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The first question is what is the volume of grain in the hopper? So before we even get to these other questions, let's see if we can answer that. So that's going to be this volume right over here of the red part, the cone made up of the grain. Pause this video and try to figure it out. Well, from previous videos, we know that volume of a cone is going to be 1 3rd times the area of the base times the height. Now we know the height is six meters, but what we need to do is figure out the area of the base. Well, how do we do that?
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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Pause this video and try to figure it out. Well, from previous videos, we know that volume of a cone is going to be 1 3rd times the area of the base times the height. Now we know the height is six meters, but what we need to do is figure out the area of the base. Well, how do we do that? Well, we'd have to figure out the radius of the base. Let's call that R right over here. And how do we figure that out?
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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Well, how do we do that? Well, we'd have to figure out the radius of the base. Let's call that R right over here. And how do we figure that out? Well, we can look at these two triangles that you can see on my screen and realize that they are similar triangles. This line is parallel to that line. This is a right angle.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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And how do we figure that out? Well, we can look at these two triangles that you can see on my screen and realize that they are similar triangles. This line is parallel to that line. This is a right angle. This is a right angle because both of these cuts of these surfaces are going to be parallel to the ground. And then this angle is going to be congruent to this angle because you could view this line as a transversal between parallel lines and these are corresponding angles. And then both triangles share this.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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This is a right angle. This is a right angle because both of these cuts of these surfaces are going to be parallel to the ground. And then this angle is going to be congruent to this angle because you could view this line as a transversal between parallel lines and these are corresponding angles. And then both triangles share this. So you have angle, angle, angle. These are similar triangles. And so we can set up a proportion here.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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And then both triangles share this. So you have angle, angle, angle. These are similar triangles. And so we can set up a proportion here. We can say the ratio between R and 10 meters, the ratio of R to 10 is equal to the ratio of six to eight, is equal to the ratio of six to eight. And then we could try to solve for R. R is going to be equal to, R is equal to multiply both sides by 10, multiply both sides by 10, and you're going to get 60 over eight. 60 over eight, eight goes into 60 seven times with four left over.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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And so we can set up a proportion here. We can say the ratio between R and 10 meters, the ratio of R to 10 is equal to the ratio of six to eight, is equal to the ratio of six to eight. And then we could try to solve for R. R is going to be equal to, R is equal to multiply both sides by 10, multiply both sides by 10, and you're going to get 60 over eight. 60 over eight, eight goes into 60 seven times with four left over. So it's seven and four eighths, or it's also 7.5. And so if you wanna know the area of the base right over here, if you wanted to know this B, it would be pi times the radius squared. So B in this case is going to be pi times 7.5.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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60 over eight, eight goes into 60 seven times with four left over. So it's seven and four eighths, or it's also 7.5. And so if you wanna know the area of the base right over here, if you wanted to know this B, it would be pi times the radius squared. So B in this case is going to be pi times 7.5. We're dealing with meters squared. And so the volume, to answer the first question, the volume is going to be one third times the area of the base, this area up here, which is pi times 7.5 meters squared times the height. So times six meters.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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So B in this case is going to be pi times 7.5. We're dealing with meters squared. And so the volume, to answer the first question, the volume is going to be one third times the area of the base, this area up here, which is pi times 7.5 meters squared times the height. So times six meters. And let's see, we could simplify this a little bit. Six divided by three, or six times one third, is just going to be equal to two. And so let me get my calculator.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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So times six meters. And let's see, we could simplify this a little bit. Six divided by three, or six times one third, is just going to be equal to two. And so let me get my calculator. They say round to the nearest tenth of a cubic meter. So we have 7.5 squared times two times pi is equal to, if we round to the nearest tenth, it's gonna be 353.4 cubic meters. So the volume is approximately 353.4 cubic meters.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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And so let me get my calculator. They say round to the nearest tenth of a cubic meter. So we have 7.5 squared times two times pi is equal to, if we round to the nearest tenth, it's gonna be 353.4 cubic meters. So the volume is approximately 353.4 cubic meters. So that's the answer to the first part right over there. And then they say how many complete boxes will the grain fill? Well, they talk about the boxes right over here.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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So the volume is approximately 353.4 cubic meters. So that's the answer to the first part right over there. And then they say how many complete boxes will the grain fill? Well, they talk about the boxes right over here. The hopper will pour the grain into boxes with dimensions of 0.5 meters by 0.5 meters by 0.4 meters. So we can imagine these boxes. They look like this.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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Well, they talk about the boxes right over here. The hopper will pour the grain into boxes with dimensions of 0.5 meters by 0.5 meters by 0.4 meters. So we can imagine these boxes. They look like this. And they are 0.5 meters by 0.5 meters by 0.4 meters. So the volume of each box is just going to be the product of these three numbers. So the volume of each box is going to be the width times the depth times the height.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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They look like this. And they are 0.5 meters by 0.5 meters by 0.4 meters. So the volume of each box is just going to be the product of these three numbers. So the volume of each box is going to be the width times the depth times the height. So 0.5 meters times 0.5 meters times 0.4 meters. And we should be able to do this in our head because five times five is 25. 25 times four is equal to 100.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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So the volume of each box is going to be the width times the depth times the height. So 0.5 meters times 0.5 meters times 0.4 meters. And we should be able to do this in our head because five times five is 25. 25 times four is equal to 100. But then we have to think we have one, two, three digits to the right of the decimal point. So one, two, three. So this is going to be 1 0.100 cubic meters.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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25 times four is equal to 100. But then we have to think we have one, two, three digits to the right of the decimal point. So one, two, three. So this is going to be 1 0.100 cubic meters. So a tenth of a cubic meter. So how many tenths of cubic meters can I fill up with this much grain? Well, it's just going to be this number divided by a tenth.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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So this is going to be 1 0.100 cubic meters. So a tenth of a cubic meter. So how many tenths of cubic meters can I fill up with this much grain? Well, it's just going to be this number divided by a tenth. Well, if you divide by a tenth, that's the same thing as multiplying by 10. And so if you multiply this by 10, you're going to get 3,534 boxes. Now, once again, let's just appreciate.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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Well, it's just going to be this number divided by a tenth. Well, if you divide by a tenth, that's the same thing as multiplying by 10. And so if you multiply this by 10, you're going to get 3,534 boxes. Now, once again, let's just appreciate. Every cubic meter, you can fill 10 of these boxes. And this is how many cubic meters we have. So if you multiply this by 10, it tells you how many boxes you fill up.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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Now, once again, let's just appreciate. Every cubic meter, you can fill 10 of these boxes. And this is how many cubic meters we have. So if you multiply this by 10, it tells you how many boxes you fill up. And one way to think about it, we've seen this in other videos, we're shifting the decimal one place over to the right to get this many boxes. And it's important to realize complete boxes because when we got to 353.4, we did round down. So we do have that amount, but we're not going to fill up another box with whatever this rounding error that we rounded down from.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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So if you multiply this by 10, it tells you how many boxes you fill up. And one way to think about it, we've seen this in other videos, we're shifting the decimal one place over to the right to get this many boxes. And it's important to realize complete boxes because when we got to 353.4, we did round down. So we do have that amount, but we're not going to fill up another box with whatever this rounding error that we rounded down from. So the last question is, to the nearest minute, how long does it take to fill the boxes? Well, this is the total volume, and we're going to fill eight cubic meters per minute. So the answer over here is going to be our total volume, it's going to be 353.4 cubic meters.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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So we do have that amount, but we're not going to fill up another box with whatever this rounding error that we rounded down from. So the last question is, to the nearest minute, how long does it take to fill the boxes? Well, this is the total volume, and we're going to fill eight cubic meters per minute. So the answer over here is going to be our total volume, it's going to be 353.4 cubic meters. And we're going to divide that by our rate, eight cubic meters per minute. And that is going to give us 353.4 divided by eight is equal to, and if we want to round to the nearest minute, 44 minutes is equal to approximately 44 minutes to fill all the boxes. And we're done.
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Applying volume of solids Solid geometry High school geometry Khan Academy.mp3
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You can't be just perpendicular by yourself. And perpendicular lines, just so you have a visualization for what perpendicular lines look like. Two lines are perpendicular if they intersect at right angles. So if this is one line right there, a perpendicular line will look like this. A perpendicular line will intersect it, but it won't just be any intersection. It will intersect at right angles. It will intersect at right angles.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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So if this is one line right there, a perpendicular line will look like this. A perpendicular line will intersect it, but it won't just be any intersection. It will intersect at right angles. It will intersect at right angles. So these two lines are perpendicular. Now, if two lines are perpendicular, if the slope of this orange line is m, so let's say its equation is y is equal to mx plus, let's say it's b1, so it's some y-intercept, then the equation of this yellow line, its slope is going to be the negative inverse of this guy. This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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It will intersect at right angles. So these two lines are perpendicular. Now, if two lines are perpendicular, if the slope of this orange line is m, so let's say its equation is y is equal to mx plus, let's say it's b1, so it's some y-intercept, then the equation of this yellow line, its slope is going to be the negative inverse of this guy. This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept. Or another way to think about it is if two lines are perpendicular, the product of their slopes is going to be negative 1. And so you could write that there. m times negative 1 over m. That's going to be, these two guys are going to cancel out, that's going to be equal to negative 1.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept. Or another way to think about it is if two lines are perpendicular, the product of their slopes is going to be negative 1. And so you could write that there. m times negative 1 over m. That's going to be, these two guys are going to cancel out, that's going to be equal to negative 1. So let's figure out the slopes of each of these lines and figure out if any of them are the negative inverse of any of the other ones. So line A, the slope is pretty easy to figure out. It's already in slope-intercept form.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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m times negative 1 over m. That's going to be, these two guys are going to cancel out, that's going to be equal to negative 1. So let's figure out the slopes of each of these lines and figure out if any of them are the negative inverse of any of the other ones. So line A, the slope is pretty easy to figure out. It's already in slope-intercept form. Its slope is 3. So line A has a slope of 3. Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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It's already in slope-intercept form. Its slope is 3. So line A has a slope of 3. Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it. So let's do line B over here. Line B, we have x plus 3y is equal to negative 21. Let's subtract x from both sides so that it ends up on the right-hand side.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it. So let's do line B over here. Line B, we have x plus 3y is equal to negative 21. Let's subtract x from both sides so that it ends up on the right-hand side. So this, we end up with 3y is equal to negative x minus 21. And now let's divide both sides of this equation by 3. And we get y is equal to negative 1 third x minus 7.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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Let's subtract x from both sides so that it ends up on the right-hand side. So this, we end up with 3y is equal to negative x minus 21. And now let's divide both sides of this equation by 3. And we get y is equal to negative 1 third x minus 7. So this character's slope is negative 1 third. So here, m is equal to negative 1 third. So we already see they are the negative inverse of each other.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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And we get y is equal to negative 1 third x minus 7. So this character's slope is negative 1 third. So here, m is equal to negative 1 third. So we already see they are the negative inverse of each other. You take the inverse of 3, it's 1 third, and then it's the negative of that. Or you take the inverse of negative 1 third, it's negative 3, and then this is the negative of that. So these two lines are definitely perpendicular.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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So we already see they are the negative inverse of each other. You take the inverse of 3, it's 1 third, and then it's the negative of that. Or you take the inverse of negative 1 third, it's negative 3, and then this is the negative of that. So these two lines are definitely perpendicular. Let's see this third line over here. So line C is 3x plus y is equal to 10. If we subtract 3x from both sides, we get y is equal to negative 3x plus 10.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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So these two lines are definitely perpendicular. Let's see this third line over here. So line C is 3x plus y is equal to 10. If we subtract 3x from both sides, we get y is equal to negative 3x plus 10. So our slope in this case is negative 3. So our slope here is equal to negative 3. Now this guy is the negative of that guy.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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If we subtract 3x from both sides, we get y is equal to negative 3x plus 10. So our slope in this case is negative 3. So our slope here is equal to negative 3. Now this guy is the negative of that guy. This guy's slope is the negative of that, but not the negative inverse. So it's not perpendicular. And this guy is the inverse of that guy, but not the negative inverse.
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Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3
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And the transformations we're going to look at are things like rotations, where you are spinning something around a point. We're gonna look at translations, where you're shifting all the points of a figure. We're gonna look at reflection, where you flip a figure over some type of a line. And we'll look at dilations, where you're essentially going to either shrink or expand some type of a figure. So with that out of the way, let's think about this question. What single transformation was applied to triangle A to get triangle B? So it looks like triangle A and triangle B, they're the same size.
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Examples recognizing transformations.mp3
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And we'll look at dilations, where you're essentially going to either shrink or expand some type of a figure. So with that out of the way, let's think about this question. What single transformation was applied to triangle A to get triangle B? So it looks like triangle A and triangle B, they're the same size. And what's really happened is that every one of these points has been shifted, or another way I could say it, they have all been translated a little bit to the right and up. And so, right like this, they have all been translated. So this right over here is clearly a translation.
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Examples recognizing transformations.mp3
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So it looks like triangle A and triangle B, they're the same size. And what's really happened is that every one of these points has been shifted, or another way I could say it, they have all been translated a little bit to the right and up. And so, right like this, they have all been translated. So this right over here is clearly a translation. Let's do another example. What single transformation was applied to get, what is applied to triangle A to get to triangle B? So if I look at this, these diagrams, this point seems to correspond with that one.
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Examples recognizing transformations.mp3
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So this right over here is clearly a translation. Let's do another example. What single transformation was applied to get, what is applied to triangle A to get to triangle B? So if I look at this, these diagrams, this point seems to correspond with that one. This one corresponds with that one. So it doesn't look like straight translation, because they would have been translated in different ways. So it's definitely not a straight translation.
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Examples recognizing transformations.mp3
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So if I look at this, these diagrams, this point seems to correspond with that one. This one corresponds with that one. So it doesn't look like straight translation, because they would have been translated in different ways. So it's definitely not a straight translation. Let's think about it. It looks like there might be a rotation here. So maybe it looks like that point went over there, that point went over there, this point went over here.
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Examples recognizing transformations.mp3
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So it's definitely not a straight translation. Let's think about it. It looks like there might be a rotation here. So maybe it looks like that point went over there, that point went over there, this point went over here. And so we could be rotating around some point right about here. And if you rotate around that point, you could get to a situation that looks like triangle B. And I don't know the exact point that we're rotating around, but this looks pretty clear, like a rotation.
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Examples recognizing transformations.mp3
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So maybe it looks like that point went over there, that point went over there, this point went over here. And so we could be rotating around some point right about here. And if you rotate around that point, you could get to a situation that looks like triangle B. And I don't know the exact point that we're rotating around, but this looks pretty clear, like a rotation. Let's do another example. What single transformation was applied to quadrilateral A to get to quadrilateral B? So let's see, it looks like this point corresponds to that point.
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Examples recognizing transformations.mp3
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And I don't know the exact point that we're rotating around, but this looks pretty clear, like a rotation. Let's do another example. What single transformation was applied to quadrilateral A to get to quadrilateral B? So let's see, it looks like this point corresponds to that point. And so, and then this point corresponds to that point, and that point corresponds to that point. So they actually look like reflections of each other. So we're to imagine some type of a mirror right over here.
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Examples recognizing transformations.mp3
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So let's see, it looks like this point corresponds to that point. And so, and then this point corresponds to that point, and that point corresponds to that point. So they actually look like reflections of each other. So we're to imagine some type of a mirror right over here. They're actually mirror images. This got flipped over the line, that got flipped over the line, and that got flipped over the line. So it's pretty clear that this right over here is a reflection.
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Examples recognizing transformations.mp3
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So we're to imagine some type of a mirror right over here. They're actually mirror images. This got flipped over the line, that got flipped over the line, and that got flipped over the line. So it's pretty clear that this right over here is a reflection. All right, let's do one more of these. What single transformation was applied to quadrilateral A to get to quadrilateral B? All right.
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Examples recognizing transformations.mp3
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So it's pretty clear that this right over here is a reflection. All right, let's do one more of these. What single transformation was applied to quadrilateral A to get to quadrilateral B? All right. So this looks like, so quadrilateral B is clearly bigger. So this is a non-rigid transformation. The distance between corresponding points looks like it has increased.
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Examples recognizing transformations.mp3
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All right. So this looks like, so quadrilateral B is clearly bigger. So this is a non-rigid transformation. The distance between corresponding points looks like it has increased. Now you might be saying, well, wouldn't that be, it looks like if you're making something bigger or smaller, that looks like a dilation. But it looks like this has been moved as well, as it has been translated. And the key here to realize is around, what is your center of dilation?
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Examples recognizing transformations.mp3
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The distance between corresponding points looks like it has increased. Now you might be saying, well, wouldn't that be, it looks like if you're making something bigger or smaller, that looks like a dilation. But it looks like this has been moved as well, as it has been translated. And the key here to realize is around, what is your center of dilation? So for example, if your center of dilation is, let's say right over here, then all of these things are going to be stretched that way. And so this point might go to there, that point might go over there, this point might go, this point might go over here, and then that point might go over here. So this is definitely a dilation, where you are, your center, where everything is expanding from, is just outside of our trapezoid A.
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Examples recognizing transformations.mp3
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And the reason why I wrote angle side angle here and angle angle side is to realize that these are equivalent, because if you have two angles, then you know what the third angle is going to be. So for example, in this case right over here, if we know that we have two pairs of angles that have the same measure, then that means that the third pair must have the same measure as well. So we'll know this as well. So if you really think about it, if you have the side between the two angles, that's equivalent to having an angle, an angle, and a side, because as long as you have two angles, the third angle is also going to have the same measure as the corresponding third angle on the other triangle. So let's just show a series of rigid transformations that can get us from ABC to DEF. So the first step you might imagine, we've already shown that if you have two segments of equal length, that they are congruent, you can have a series of rigid transformations that maps one onto the other. So what I want to do is map segment AC onto DF.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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So if you really think about it, if you have the side between the two angles, that's equivalent to having an angle, an angle, and a side, because as long as you have two angles, the third angle is also going to have the same measure as the corresponding third angle on the other triangle. So let's just show a series of rigid transformations that can get us from ABC to DEF. So the first step you might imagine, we've already shown that if you have two segments of equal length, that they are congruent, you can have a series of rigid transformations that maps one onto the other. So what I want to do is map segment AC onto DF. And the way that I could do that is I could translate point A to be on top of point D. So then I'll call this A prime. And then when I do that, this segment AC is going to look something like this. I'm just sketching it right now.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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So what I want to do is map segment AC onto DF. And the way that I could do that is I could translate point A to be on top of point D. So then I'll call this A prime. And then when I do that, this segment AC is going to look something like this. I'm just sketching it right now. It's going to be in that direction. But then, and the whole, the rest of the triangle is going to come with it. So let's see, the rest of that orange side, side AB is going to look something like that.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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I'm just sketching it right now. It's going to be in that direction. But then, and the whole, the rest of the triangle is going to come with it. So let's see, the rest of that orange side, side AB is going to look something like that. But then we could do another rigid transformation, which is rotate about point D or point A prime. They're the same point now. So that point C coincides with point F. And so just like that, you would have two rigid transformations that get us, that map AC onto DF.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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So let's see, the rest of that orange side, side AB is going to look something like that. But then we could do another rigid transformation, which is rotate about point D or point A prime. They're the same point now. So that point C coincides with point F. And so just like that, you would have two rigid transformations that get us, that map AC onto DF. And so A prime, where A is mapped, is now equal to D, and F is now equal to C prime. But the question is, where does point B now sit? And the realization here is that angle measures are preserved.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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So that point C coincides with point F. And so just like that, you would have two rigid transformations that get us, that map AC onto DF. And so A prime, where A is mapped, is now equal to D, and F is now equal to C prime. But the question is, where does point B now sit? And the realization here is that angle measures are preserved. And since angle measures are preserved, we are either going to have a situation where this angle, let's see, this angle is angle CAB gets preserved. So then it would be C prime, A prime, and then B prime would have to sit someplace on this ray for if we're gonna preserve the measure of angle CAB, B prime is going to sit someplace along that ray because an angle is defined by two rays that intersect at the vertex or start at the vertex. And because this angle is preserved, that's the angle that is formed by these two rays.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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And the realization here is that angle measures are preserved. And since angle measures are preserved, we are either going to have a situation where this angle, let's see, this angle is angle CAB gets preserved. So then it would be C prime, A prime, and then B prime would have to sit someplace on this ray for if we're gonna preserve the measure of angle CAB, B prime is going to sit someplace along that ray because an angle is defined by two rays that intersect at the vertex or start at the vertex. And because this angle is preserved, that's the angle that is formed by these two rays. You could say ray CA and ray CB. We know that B prime also has to sit someplace on this ray as well. So B prime also has to sit someplace on this ray.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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And because this angle is preserved, that's the angle that is formed by these two rays. You could say ray CA and ray CB. We know that B prime also has to sit someplace on this ray as well. So B prime also has to sit someplace on this ray. And I think you see where this is going. If B prime, because these two angles are preserved, because this angle and this angle are preserved, have to sit someplace on both of these rays, they intersect at one point, this point right over here that coincides with point E. So this is where B prime would be. So that's one scenario in which case we've shown that you can get a series of rigid transformations from this triangle to this triangle.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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So B prime also has to sit someplace on this ray. And I think you see where this is going. If B prime, because these two angles are preserved, because this angle and this angle are preserved, have to sit someplace on both of these rays, they intersect at one point, this point right over here that coincides with point E. So this is where B prime would be. So that's one scenario in which case we've shown that you can get a series of rigid transformations from this triangle to this triangle. But there's another one. There is a circumstance where the angles get preserved, but instead of being on, instead of the angles being on the, I guess you could say the bottom right side of this blue line, you could imagine the angles get preserved such that they are on the other side. So the angles get preserved so that they are on the other side of that blue line.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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So that's one scenario in which case we've shown that you can get a series of rigid transformations from this triangle to this triangle. But there's another one. There is a circumstance where the angles get preserved, but instead of being on, instead of the angles being on the, I guess you could say the bottom right side of this blue line, you could imagine the angles get preserved such that they are on the other side. So the angles get preserved so that they are on the other side of that blue line. And then the question is, in that situation, where would B prime end up? Well, actually, let me draw this a little bit, let me do this a little bit more exact. Let me replicate these angles.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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So the angles get preserved so that they are on the other side of that blue line. And then the question is, in that situation, where would B prime end up? Well, actually, let me draw this a little bit, let me do this a little bit more exact. Let me replicate these angles. So I'm going to draw an arc like this, an arc like this, and then I'll measure this distance. It's just like this. We've done this in other videos when we're trying to replicate angles.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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Let me replicate these angles. So I'm going to draw an arc like this, an arc like this, and then I'll measure this distance. It's just like this. We've done this in other videos when we're trying to replicate angles. So it's like that far. And so let me draw that on this point right over here, this far. So if the angles are on that side of line, I guess we could say DF or A prime, C prime, we know that B prime would have to sit someplace on this ray.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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We've done this in other videos when we're trying to replicate angles. So it's like that far. And so let me draw that on this point right over here, this far. So if the angles are on that side of line, I guess we could say DF or A prime, C prime, we know that B prime would have to sit someplace on this ray. So let me draw that as neatly as I can. Someplace on this ray. And it would have to sit someplace on the ray formed by the other angle.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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So if the angles are on that side of line, I guess we could say DF or A prime, C prime, we know that B prime would have to sit someplace on this ray. So let me draw that as neatly as I can. Someplace on this ray. And it would have to sit someplace on the ray formed by the other angle. So let me see if I can draw that as neatly as possible. So let me make a, make a arc like this. I probably did that a little bit bigger than I need to, but hopefully it serves our purposes.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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And it would have to sit someplace on the ray formed by the other angle. So let me see if I can draw that as neatly as possible. So let me make a, make a arc like this. I probably did that a little bit bigger than I need to, but hopefully it serves our purposes. I measured this distance right over here. If I measure that distance over here, it would get us right over there. So B prime either sits on this ray, or it could sit, or and it has to sit, I should really say, on this ray.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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I probably did that a little bit bigger than I need to, but hopefully it serves our purposes. I measured this distance right over here. If I measure that distance over here, it would get us right over there. So B prime either sits on this ray, or it could sit, or and it has to sit, I should really say, on this ray. It goes through this point and this point. And it has to sit on this ray. And you can see where these two rays intersect is right over there.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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So B prime either sits on this ray, or it could sit, or and it has to sit, I should really say, on this ray. It goes through this point and this point. And it has to sit on this ray. And you can see where these two rays intersect is right over there. So the other scenario is if the angles get preserved in a way that they're on the other side of that blue line, well then B prime is there. And then we could just add one more rigid transformation to our series of rigid transformations, which is essentially, or is, a reflection across line DF, or A prime, C prime. Why will that work to map B prime onto E?
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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And you can see where these two rays intersect is right over there. So the other scenario is if the angles get preserved in a way that they're on the other side of that blue line, well then B prime is there. And then we could just add one more rigid transformation to our series of rigid transformations, which is essentially, or is, a reflection across line DF, or A prime, C prime. Why will that work to map B prime onto E? Well, because reflection is also a rigid transformation, so angles are preserved. And so as this angle gets flipped over, it's preserved. As this angle gets flipped over, the measure of it, I should say, is preserved.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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Why will that work to map B prime onto E? Well, because reflection is also a rigid transformation, so angles are preserved. And so as this angle gets flipped over, it's preserved. As this angle gets flipped over, the measure of it, I should say, is preserved. And so that means we'll go to that first case where then these rays would be flipped onto these rays, and B prime would have to sit on that intersection. And there you have it. If you have two angles, and if you have two angles, you're gonna know the third.
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Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3
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So you have parallel, you have perpendicular, perpendicular, and of course you have lines that are neither parallel nor perpendicular. And just as a bit of a review, if you've never seen this before, parallel lines, they never intersect. So let me draw some axes. So if those are my coordinate axes right there, that's my x-axis, that is my y-axis. If this is a line that I'm drawing in magenta, a parallel line might look something like this. It's not the exact same line, but they have the exact same slope. If this moves a certain amount, if this change in y over change in x is a certain amount, this change in y over change in x is the same amount, and that's why they never intersect.
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Parallel & perpendicular lines from graph.mp3
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So if those are my coordinate axes right there, that's my x-axis, that is my y-axis. If this is a line that I'm drawing in magenta, a parallel line might look something like this. It's not the exact same line, but they have the exact same slope. If this moves a certain amount, if this change in y over change in x is a certain amount, this change in y over change in x is the same amount, and that's why they never intersect. So they have the same slope. Parallel lines have the same slope. Perpendicular lines, depending on how you want to view it, they're kind of the opposite.
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Parallel & perpendicular lines from graph.mp3
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If this moves a certain amount, if this change in y over change in x is a certain amount, this change in y over change in x is the same amount, and that's why they never intersect. So they have the same slope. Parallel lines have the same slope. Perpendicular lines, depending on how you want to view it, they're kind of the opposite. Let's say that this is some line. A line that is perpendicular to that will not only intersect the line, it will intersect it at a right angle, at a 90 degree angle. I'm not going to prove it for you here, I actually prove it in the linear algebra playlist.
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Parallel & perpendicular lines from graph.mp3
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Perpendicular lines, depending on how you want to view it, they're kind of the opposite. Let's say that this is some line. A line that is perpendicular to that will not only intersect the line, it will intersect it at a right angle, at a 90 degree angle. I'm not going to prove it for you here, I actually prove it in the linear algebra playlist. But a perpendicular line's slope, so let's say that this one right here, let's say that yellow line has a slope of m, then this orange line that's perpendicular to the yellow line is going to have a slope of negative 1 over m. Their slopes are going to be the negative inverse of each other. Now, given this information, let's look at a bunch of lines and figure out if they're parallel, if they're perpendicular, or if they are neither. To do that, we just have to keep looking at the slopes.
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Parallel & perpendicular lines from graph.mp3
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I'm not going to prove it for you here, I actually prove it in the linear algebra playlist. But a perpendicular line's slope, so let's say that this one right here, let's say that yellow line has a slope of m, then this orange line that's perpendicular to the yellow line is going to have a slope of negative 1 over m. Their slopes are going to be the negative inverse of each other. Now, given this information, let's look at a bunch of lines and figure out if they're parallel, if they're perpendicular, or if they are neither. To do that, we just have to keep looking at the slopes. So let's see. They say one line passes through the points 4, negative 3, and negative 8, 0. Another line passes through the points negative 1, negative 1, and negative 2, 6.
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Parallel & perpendicular lines from graph.mp3
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To do that, we just have to keep looking at the slopes. So let's see. They say one line passes through the points 4, negative 3, and negative 8, 0. Another line passes through the points negative 1, negative 1, and negative 2, 6. Let's figure out the slopes of each of these lines. I'll first do this one in pink. This slope right here, so line 1, I'll call it slope 1.
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Parallel & perpendicular lines from graph.mp3
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Another line passes through the points negative 1, negative 1, and negative 2, 6. Let's figure out the slopes of each of these lines. I'll first do this one in pink. This slope right here, so line 1, I'll call it slope 1. Slope 1 is, let's just say it is, well, let's take this as the finishing point. Negative 3 minus 0, remember, change in y, over 4 minus negative 8. So this is equal to negative 3 over, this is the same thing as 4 plus 8, negative 3 over 12, which is equal to negative 1 fourth, divide the numerator and denominator by 3.
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Parallel & perpendicular lines from graph.mp3
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This slope right here, so line 1, I'll call it slope 1. Slope 1 is, let's just say it is, well, let's take this as the finishing point. Negative 3 minus 0, remember, change in y, over 4 minus negative 8. So this is equal to negative 3 over, this is the same thing as 4 plus 8, negative 3 over 12, which is equal to negative 1 fourth, divide the numerator and denominator by 3. That's this line, that's the first line. Now what about the second line? The second line, the slope for that second line is, well, let's take here, negative 1 minus 6 over negative 1 minus negative 2, is equal to negative 1 minus 6 is negative 7, over negative 1 minus negative 2, that's the same thing as negative 1 plus 2.
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Parallel & perpendicular lines from graph.mp3
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So this is equal to negative 3 over, this is the same thing as 4 plus 8, negative 3 over 12, which is equal to negative 1 fourth, divide the numerator and denominator by 3. That's this line, that's the first line. Now what about the second line? The second line, the slope for that second line is, well, let's take here, negative 1 minus 6 over negative 1 minus negative 2, is equal to negative 1 minus 6 is negative 7, over negative 1 minus negative 2, that's the same thing as negative 1 plus 2. Well, that's just 1. So the slope here is negative 7. So here their slopes are neither equal, so they're not parallel, nor are they the negative inverse of each other.
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Parallel & perpendicular lines from graph.mp3
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The second line, the slope for that second line is, well, let's take here, negative 1 minus 6 over negative 1 minus negative 2, is equal to negative 1 minus 6 is negative 7, over negative 1 minus negative 2, that's the same thing as negative 1 plus 2. Well, that's just 1. So the slope here is negative 7. So here their slopes are neither equal, so they're not parallel, nor are they the negative inverse of each other. So this is neither. This is neither parallel nor perpendicular. So these two lines, they intersect, but they're not going to intersect at a 90 degree angle.
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Parallel & perpendicular lines from graph.mp3
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So here their slopes are neither equal, so they're not parallel, nor are they the negative inverse of each other. So this is neither. This is neither parallel nor perpendicular. So these two lines, they intersect, but they're not going to intersect at a 90 degree angle. Let's do a couple more of these. So I have here, once again, one line passing through these points, and then another line passing through these points. So let's just look at their slopes.
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Parallel & perpendicular lines from graph.mp3
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So these two lines, they intersect, but they're not going to intersect at a 90 degree angle. Let's do a couple more of these. So I have here, once again, one line passing through these points, and then another line passing through these points. So let's just look at their slopes. So this one in green, what's the slope? The slope of the green one, I'll call that the first line. We could say, let's see, change in y, so we could do negative 2 minus 14 over, I did negative 2 first, so I'll do 1 first, over 1 minus negative 3.
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Parallel & perpendicular lines from graph.mp3
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So let's just look at their slopes. So this one in green, what's the slope? The slope of the green one, I'll call that the first line. We could say, let's see, change in y, so we could do negative 2 minus 14 over, I did negative 2 first, so I'll do 1 first, over 1 minus negative 3. So negative 2 minus 14 is negative 16. 1 minus negative 3, the same thing as 1 plus 3, that's over 4. So this is negative 4.
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Parallel & perpendicular lines from graph.mp3
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We could say, let's see, change in y, so we could do negative 2 minus 14 over, I did negative 2 first, so I'll do 1 first, over 1 minus negative 3. So negative 2 minus 14 is negative 16. 1 minus negative 3, the same thing as 1 plus 3, that's over 4. So this is negative 4. Now what's the slope of that second line right there? So we have the slope of that second line, let's say 5 minus negative 3, that's our change in y, over negative 2 minus 0. So this is equal to 5 minus negative 3, that's the same thing as 5 plus 3, that's 8.
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Parallel & perpendicular lines from graph.mp3
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So this is negative 4. Now what's the slope of that second line right there? So we have the slope of that second line, let's say 5 minus negative 3, that's our change in y, over negative 2 minus 0. So this is equal to 5 minus negative 3, that's the same thing as 5 plus 3, that's 8. And then negative 2 minus 0 is negative 2. So this is also equal to negative 4. So these two lines are parallel.
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Parallel & perpendicular lines from graph.mp3
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So this is equal to 5 minus negative 3, that's the same thing as 5 plus 3, that's 8. And then negative 2 minus 0 is negative 2. So this is also equal to negative 4. So these two lines are parallel. These two lines are parallel. They have the exact same slope. And I encourage you to find the equations of both of these lines and graph both of these lines and verify for yourself that they are indeed parallel.
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Parallel & perpendicular lines from graph.mp3
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