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We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it, just knowing what you know about the sums of the measures of the angles inside of a triangle and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself, because I'm about to give you the solution. So the first thing you might say, and this is a general way to think about a lot of these problems, where they give you some angles and you have to figure out some other angles based on the sum of angles in a triangle equaling 180, or this one doesn't have parallel lines on it, but you might see some with parallel lines and supplementary lines and complementary lines, is to just fill in everything that you can figure out. And one way or another, you'll probably be able to figure out what this question mark is.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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And so you might want to give a go at it, just knowing what you know about the sums of the measures of the angles inside of a triangle and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself, because I'm about to give you the solution. So the first thing you might say, and this is a general way to think about a lot of these problems, where they give you some angles and you have to figure out some other angles based on the sum of angles in a triangle equaling 180, or this one doesn't have parallel lines on it, but you might see some with parallel lines and supplementary lines and complementary lines, is to just fill in everything that you can figure out. And one way or another, you'll probably be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given two of the angles.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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And one way or another, you'll probably be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given two of the angles. And if you have two of the angles in a triangle, you can always figure out the third angle, because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say x plus, what is this, 114 is equal to 180 degrees.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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And on this triangle on the left, we're given two of the angles. And if you have two of the angles in a triangle, you can always figure out the third angle, because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say x plus, what is this, 114 is equal to 180 degrees. We can subtract 114 from both sides of this equation. And we get x is equal to 180 minus 114. So 80 minus 14, 80 minus 10 would be 70, minus another 4 is 66.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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Or we could say x plus, what is this, 114 is equal to 180 degrees. We can subtract 114 from both sides of this equation. And we get x is equal to 180 minus 114. So 80 minus 14, 80 minus 10 would be 70, minus another 4 is 66. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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So 80 minus 14, 80 minus 10 would be 70, minus another 4 is 66. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. So x is equal to 66 degrees. Well, if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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So let me write it like this. So x is equal to 66 degrees. Well, if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides. And we get y is equal to, these cancel out, 180 minus 66 is 114. And that number might look a little familiar to you.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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So this is 66. And so we can subtract 66 from both sides. And we get y is equal to, these cancel out, 180 minus 66 is 114. And that number might look a little familiar to you. Notice this 114 was the exact same sum of these two angles over here. And that's actually a general idea. And I'll do it on the side here just to prove it to you.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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And that number might look a little familiar to you. Notice this 114 was the exact same sum of these two angles over here. And that's actually a general idea. And I'll do it on the side here just to prove it to you. If I have, let's say that these two angles, let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle, so in this example, y is an exterior angle.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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And I'll do it on the side here just to prove it to you. If I have, let's say that these two angles, let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle, so in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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And then this angle, which is considered to be an exterior angle, so in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You can subtract 180 from both sides. You can add a plus b to both sides.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You can subtract 180 from both sides. You can add a plus b to both sides. So plus a plus b, running out of space on the right-hand side. And then you're left with, these cancel out. On the left-hand side, you're left with y.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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You can add a plus b to both sides. So plus a plus b, running out of space on the right-hand side. And then you're left with, these cancel out. On the left-hand side, you're left with y. On the right-hand side, is equal to a plus b. So this is just a general property. You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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On the left-hand side, you're left with y. On the right-hand side, is equal to a plus b. So this is just a general property. You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees. And then you have the supplementary angles right over here. Or you could just say, look, if I have the exterior angles right over here, it's equal to the sum of the remote interior angles. That's just a little terminology you'd see there.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees. And then you have the supplementary angles right over here. Or you could just say, look, if I have the exterior angles right over here, it's equal to the sum of the remote interior angles. That's just a little terminology you'd see there. So y is equal to a plus b. 114 degrees, we've already shown to ourselves, is equal to 64 plus 50 degrees. But anyway, regardless of how we do it, if we just reason it out step by step, or if we just knew this property from the get-go, if we know that y is equal to 114 degrees, and I like to reason it out every time just to make sure I'm not jumping to conclusions.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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That's just a little terminology you'd see there. So y is equal to a plus b. 114 degrees, we've already shown to ourselves, is equal to 64 plus 50 degrees. But anyway, regardless of how we do it, if we just reason it out step by step, or if we just knew this property from the get-go, if we know that y is equal to 114 degrees, and I like to reason it out every time just to make sure I'm not jumping to conclusions. So if y is 114 degrees, now we know this angle. We were given this angle in the beginning. Now we just have to figure out this third angle in this triangle.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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But anyway, regardless of how we do it, if we just reason it out step by step, or if we just knew this property from the get-go, if we know that y is equal to 114 degrees, and I like to reason it out every time just to make sure I'm not jumping to conclusions. So if y is 114 degrees, now we know this angle. We were given this angle in the beginning. Now we just have to figure out this third angle in this triangle. So if we call this z, if we call this question mark is equal to z, we know that z plus 114 plus 31 is equal to 180 degrees. The sums of the measures of the angle inside of a triangle add up to 180 degrees. That's the only property we're using in this step.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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Now we just have to figure out this third angle in this triangle. So if we call this z, if we call this question mark is equal to z, we know that z plus 114 plus 31 is equal to 180 degrees. The sums of the measures of the angle inside of a triangle add up to 180 degrees. That's the only property we're using in this step. So we get z plus, what is this? 145 is equal to 180. Did I do that right?
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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That's the only property we're using in this step. So we get z plus, what is this? 145 is equal to 180. Did I do that right? We have a 15 and then a 30. Yep, 145 is equal to 180. Subtract 145 from both sides of this equation, and we are left with z is equal to 80 minus 45 is equal to 35.
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Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3
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We're asked which of these lines are parallel. So they give us three equations of three different lines. And if they're parallel, then they have to have the same slope. So all we have to do over here is figure out the slopes of each of these lines. And if any of them are equal, they're parallel. So let's do line A. Line A, it's 2y is equal to 12x plus 10.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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So all we have to do over here is figure out the slopes of each of these lines. And if any of them are equal, they're parallel. So let's do line A. Line A, it's 2y is equal to 12x plus 10. We're almost in slope-intercept form. We can just divide both sides of this equation by 2. Let's divide all the terms by 2.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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Line A, it's 2y is equal to 12x plus 10. We're almost in slope-intercept form. We can just divide both sides of this equation by 2. Let's divide all the terms by 2. We get y is equal to 6x. 12 divided by 2. 6x plus 5.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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Let's divide all the terms by 2. We get y is equal to 6x. 12 divided by 2. 6x plus 5. So our slope in this case, we have it in slope-intercept form, our slope in this case is equal to 6. Let's try line B. Line B is y is equal to 6.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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6x plus 5. So our slope in this case, we have it in slope-intercept form, our slope in this case is equal to 6. Let's try line B. Line B is y is equal to 6. Now this might be, you might say, hey, this is a bizarre character. How do I get this into slope-intercept form? Where is the x?
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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Line B is y is equal to 6. Now this might be, you might say, hey, this is a bizarre character. How do I get this into slope-intercept form? Where is the x? And my answer to you is that it already is in slope-intercept form. I could just rewrite it as y is equal to 0x plus 6. The x term is being multiplied by 0, because the slope here is 0. y is going to be equal to 6 no matter how much you change x.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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Where is the x? And my answer to you is that it already is in slope-intercept form. I could just rewrite it as y is equal to 0x plus 6. The x term is being multiplied by 0, because the slope here is 0. y is going to be equal to 6 no matter how much you change x. Change in y is always going to be 0. It's always going to be 6. So here, our slope is 0.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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The x term is being multiplied by 0, because the slope here is 0. y is going to be equal to 6 no matter how much you change x. Change in y is always going to be 0. It's always going to be 6. So here, our slope is 0. Our slope is equal to 0. So these two lines are definitely not parallel. They have different slopes.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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So here, our slope is 0. Our slope is equal to 0. So these two lines are definitely not parallel. They have different slopes. Let's try line C. I'll do it down here. So that's y minus 2 is equal to 6 times x plus 2. And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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They have different slopes. Let's try line C. I'll do it down here. So that's y minus 2 is equal to 6 times x plus 2. And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2. So it's a point negative 2, 2 is being represented here, because you're subtracting the points. And the slope is 6. So we already know that the slope is equal to 6.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2. So it's a point negative 2, 2 is being represented here, because you're subtracting the points. And the slope is 6. So we already know that the slope is equal to 6. And sometimes people are more comfortable with slope-intercept form. So let's put it in slope-intercept form just to confirm that if we put it in this form, the slope will still be equal to 6. So if we distribute this 6, we get y minus 2 is equal to 6 times x, 6x plus 6 times 2 is 12.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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So we already know that the slope is equal to 6. And sometimes people are more comfortable with slope-intercept form. So let's put it in slope-intercept form just to confirm that if we put it in this form, the slope will still be equal to 6. So if we distribute this 6, we get y minus 2 is equal to 6 times x, 6x plus 6 times 2 is 12. And then if you add this 2, if you add 2 to both sides of the equation, you get y, because these guys cancel out, is equal to 6x plus 14. So you see once again, the slope is 6. So line A and line C have the same slope.
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Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3
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A circle with area 81 pi has a sector with a 350 degree central angle. So this whole sector right over here that's shaded in this kind of pale orange yellowish color, that has a 350 degree central angle. So you see the central angle is a very large angle. It's going all the way around, all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector, area of sector, the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is.
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Area of a sector given a central angle Circles Geometry Khan Academy.mp3
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It's going all the way around, all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector, area of sector, the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle, over 360. So the area of the sector over the total area is equal to the degrees in the central angle over the total degrees in a circle.
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Area of a sector given a central angle Circles Geometry Khan Academy.mp3
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And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle, over 360. So the area of the sector over the total area is equal to the degrees in the central angle over the total degrees in a circle. And then we just can solve for area of the sector by multiplying both sides by 81 pi. 81 pi, 81 pi. So these cancel out.
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Area of a sector given a central angle Circles Geometry Khan Academy.mp3
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So the area of the sector over the total area is equal to the degrees in the central angle over the total degrees in a circle. And then we just can solve for area of the sector by multiplying both sides by 81 pi. 81 pi, 81 pi. So these cancel out. 350 divided by 360 is 35 over 36. And so our area, our sector area, is equal to, see in the numerator, we have 35 times, instead of 81, I'm just going to write that as, let's see, that's going to be 9 times 9 pi. And the denominator, I have 36.
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Area of a sector given a central angle Circles Geometry Khan Academy.mp3
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So these cancel out. 350 divided by 360 is 35 over 36. And so our area, our sector area, is equal to, see in the numerator, we have 35 times, instead of 81, I'm just going to write that as, let's see, that's going to be 9 times 9 pi. And the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9. And so we are left with 35 times 9.
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Area of a sector given a central angle Circles Geometry Khan Academy.mp3
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And the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9. And so we are left with 35 times 9. And neither of these are divisible by 4. So that's about as simplified as we can get it. So let's think about what 35 times 9 is.
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Area of a sector given a central angle Circles Geometry Khan Academy.mp3
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And so we are left with 35 times 9. And neither of these are divisible by 4. So that's about as simplified as we can get it. So let's think about what 35 times 9 is. 35 times 9. It's going to be 350 minus 35, which would be 315, I guess. 315.
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Area of a sector given a central angle Circles Geometry Khan Academy.mp3
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So let's think about what 35 times 9 is. 35 times 9. It's going to be 350 minus 35, which would be 315, I guess. 315. Did I do that right? Yeah, it's going to be 270 plus 45, which is 315 pi over 4. 315 pi over 4 is the area of the sector.
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Area of a sector given a central angle Circles Geometry Khan Academy.mp3
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Let's see what they expect from us if we want to add a reflection. So if I click on this, it says reflection over the line from, and then we have two coordinate pairs. So they want us to define the line that we're going to reflect over with two points on that line. So let's see if we can do that. And to do that, I think I need to write something down. So let me get my scratch pad out, and I copied and pasted the same diagram. And the line of reflection, one way to think about it, we want to map point E to this point right over here.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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So let's see if we can do that. And to do that, I think I need to write something down. So let me get my scratch pad out, and I copied and pasted the same diagram. And the line of reflection, one way to think about it, we want to map point E to this point right over here. We want to map point M to this point over here. And so between any point and its corresponding point on the image after the reflection, these should be equidistant from the line of reflection. This and this should be equidistant from the line of reflection.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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And the line of reflection, one way to think about it, we want to map point E to this point right over here. We want to map point M to this point over here. And so between any point and its corresponding point on the image after the reflection, these should be equidistant from the line of reflection. This and this should be equidistant from the line of reflection. This and this should be, E and this point should be equidistant from the line of reflection. Or another way of thinking about it, that line of reflection should contain the midpoint between these two magenta points, and it should contain the midpoint between these two deep navy blue points. So let's just calculate the midpoints.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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This and this should be equidistant from the line of reflection. This and this should be, E and this point should be equidistant from the line of reflection. Or another way of thinking about it, that line of reflection should contain the midpoint between these two magenta points, and it should contain the midpoint between these two deep navy blue points. So let's just calculate the midpoints. So we could do that with a little bit of mathematics. The coordinates for E right over here, that is, let's see, that is X equals negative four, Y is equal to negative four. And the coordinates for the corresponding point to E in the image, this is X is equal to two, X is equal to two, and Y is equal to negative six.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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So let's just calculate the midpoints. So we could do that with a little bit of mathematics. The coordinates for E right over here, that is, let's see, that is X equals negative four, Y is equal to negative four. And the coordinates for the corresponding point to E in the image, this is X is equal to two, X is equal to two, and Y is equal to negative six. So what's the midpoint between negative four, negative four, and two comma negative six? Well, you just have to take the average of the Xs and take the average of the Ys. Let me do that, actually I'll do it over here.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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And the coordinates for the corresponding point to E in the image, this is X is equal to two, X is equal to two, and Y is equal to negative six. So what's the midpoint between negative four, negative four, and two comma negative six? Well, you just have to take the average of the Xs and take the average of the Ys. Let me do that, actually I'll do it over here. So if I take the average of the Xs, it's going to be negative four, negative four, plus two, plus two, over two, that's the average of the Xs. And then the average of the Ys is gonna be negative four plus negative six over two. Negative four plus negative six over two.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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Let me do that, actually I'll do it over here. So if I take the average of the Xs, it's going to be negative four, negative four, plus two, plus two, over two, that's the average of the Xs. And then the average of the Ys is gonna be negative four plus negative six over two. Negative four plus negative six over two. And then, close the parentheses. Let's see, negative four plus two is negative two. Divided by two is negative one, so it's gonna be negative one comma, negative four plus negative six, that's the same thing as negative four minus six, which is gonna be negative 10, divided by two is negative five.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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Negative four plus negative six over two. And then, close the parentheses. Let's see, negative four plus two is negative two. Divided by two is negative one, so it's gonna be negative one comma, negative four plus negative six, that's the same thing as negative four minus six, which is gonna be negative 10, divided by two is negative five. Is, let me do that in a blue color so you see where it came from, is going to be negative five. So there you have it. That's going to be the midpoint between E and the corresponding point on its image.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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Divided by two is negative one, so it's gonna be negative one comma, negative four plus negative six, that's the same thing as negative four minus six, which is gonna be negative 10, divided by two is negative five. Is, let me do that in a blue color so you see where it came from, is going to be negative five. So there you have it. That's going to be the midpoint between E and the corresponding point on its image. So let's see if I can plot that. So this is going to be, this point right over here is going to be negative one comma negative five. So X is negative one, Y is negative five.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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That's going to be the midpoint between E and the corresponding point on its image. So let's see if I can plot that. So this is going to be, this point right over here is going to be negative one comma negative five. So X is negative one, Y is negative five. So it's this point right over here. And it does indeed look like the midpoint. It looks like it's equidistant between E and this point right over here.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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So X is negative one, Y is negative five. So it's this point right over here. And it does indeed look like the midpoint. It looks like it's equidistant between E and this point right over here. And so this should sit on the line of reflection. So now let's find the midpoint between M and this point right over here. The coordinates of M are X is negative five and Y is equal to three.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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It looks like it's equidistant between E and this point right over here. And so this should sit on the line of reflection. So now let's find the midpoint between M and this point right over here. The coordinates of M are X is negative five and Y is equal to three. The coordinates here are X is equal to seven and Y is equal to negative one. So the midpoint, the X coordinate of the midpoint is going to be the average of the X's here. So let's see, it's going to be negative five plus seven over two.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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The coordinates of M are X is negative five and Y is equal to three. The coordinates here are X is equal to seven and Y is equal to negative one. So the midpoint, the X coordinate of the midpoint is going to be the average of the X's here. So let's see, it's going to be negative five plus seven over two. And the Y coordinate of the midpoint is going to be the average of the Y coordinates. So three plus negative one over two. Let's see, negative five plus seven is positive two over two is one.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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So let's see, it's going to be negative five plus seven over two. And the Y coordinate of the midpoint is going to be the average of the Y coordinates. So three plus negative one over two. Let's see, negative five plus seven is positive two over two is one. Three minus one, three plus negative one, that's positive two over two is one. So the point one comma one is the midpoint between these two. So one comma one, just like that.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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Let's see, negative five plus seven is positive two over two is one. Three minus one, three plus negative one, that's positive two over two is one. So the point one comma one is the midpoint between these two. So one comma one, just like that. So the line of reflection is going to contain these two points and two points define a line. In fact, I could, let me draw the line of reflection just because we did all of this work. The line of reflection is going to look something like, is going to look something like, I want to draw this a little bit straighter than that.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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So one comma one, just like that. So the line of reflection is going to contain these two points and two points define a line. In fact, I could, let me draw the line of reflection just because we did all of this work. The line of reflection is going to look something like, is going to look something like, I want to draw this a little bit straighter than that. It's going to look something like, it's going to look something like, something like this. And this makes sense that this is a line of reflection. I missed that magenta point a little bit.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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The line of reflection is going to look something like, is going to look something like, I want to draw this a little bit straighter than that. It's going to look something like, it's going to look something like, something like this. And this makes sense that this is a line of reflection. I missed that magenta point a little bit. So let me go through the magenta point. Okay, there you go. This makes sense that this is a line of reflection because you see that every, you pick an arbitrary point on segment ME, say that point, and if you reflect it over this line, it's this, this is its shortest distance from the line.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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I missed that magenta point a little bit. So let me go through the magenta point. Okay, there you go. This makes sense that this is a line of reflection because you see that every, you pick an arbitrary point on segment ME, say that point, and if you reflect it over this line, it's this, this is its shortest distance from the line. You just go onto the other side of the line and equal distance and you get to its corresponding point on the image. So it makes a lot of sense that these are mirror images if this is kind of the mirror here. You can imagine that this is, you know, this is kind of the surface of the water if you're looking at it at an angle.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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This makes sense that this is a line of reflection because you see that every, you pick an arbitrary point on segment ME, say that point, and if you reflect it over this line, it's this, this is its shortest distance from the line. You just go onto the other side of the line and equal distance and you get to its corresponding point on the image. So it makes a lot of sense that these are mirror images if this is kind of the mirror here. You can imagine that this is, you know, this is kind of the surface of the water if you're looking at it at an angle. I don't know if that helps you or not. But anyway, we found two points. We found two points that define that line of reflection.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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You can imagine that this is, you know, this is kind of the surface of the water if you're looking at it at an angle. I don't know if that helps you or not. But anyway, we found two points. We found two points that define that line of reflection. So now let's use the tool to type them in. One is negative one, negative five. The other one is one comma one.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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We found two points that define that line of reflection. So now let's use the tool to type them in. One is negative one, negative five. The other one is one comma one. So let me see if I can remember that. I have a bad memory. But so one is negative one comma negative five.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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The other one is one comma one. So let me see if I can remember that. I have a bad memory. But so one is negative one comma negative five. And then the other one is one comma one. And we see it worked. We see it worked.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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But so one is negative one comma negative five. And then the other one is one comma one. And we see it worked. We see it worked. By when I did that, it actually made the reflection happen. And notice, it completely went from this point and now our blue is over the image that we wanted to get to. So we are done.
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Points on line of reflection Transformations Geometry Khan Academy.mp3
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And then on top of that, on top of that we have what you could call a right pyramid, where the height of this right pyramid, so if you start at the center of its base right over here, and you go to the top, this height right over here is one unit. And this hasn't been drawn completely to scale, and kind of the perspective skews it a little bit, but our goal here, our goal here is to figure out what is the length, what is the length of one of these, of one of these edges right over here? So either that one, or this one right over here. What is that length? And we will call that, we will call that x. And so I encourage you to pause this video, and try to think about it on your own. Remember, this is a right, this is a right pyramid, so what that tells us is that this red line, that's one unit long, it is perpendicular, it is perpendicular to this entire plane.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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What is that length? And we will call that, we will call that x. And so I encourage you to pause this video, and try to think about it on your own. Remember, this is a right, this is a right pyramid, so what that tells us is that this red line, that's one unit long, it is perpendicular, it is perpendicular to this entire plane. It's perpendicular to the top of the rectangular, of the rectangular prism. So with that in mind, I encourage you to pause the video, and see if you can figure it out. And I will give you a hint.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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Remember, this is a right, this is a right pyramid, so what that tells us is that this red line, that's one unit long, it is perpendicular, it is perpendicular to this entire plane. It's perpendicular to the top of the rectangular, of the rectangular prism. So with that in mind, I encourage you to pause the video, and see if you can figure it out. And I will give you a hint. You will have to use the Pythagorean theorem, maybe more than once. All right, so I'm assuming you've at least given it a shot. So let's work through it together.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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And I will give you a hint. You will have to use the Pythagorean theorem, maybe more than once. All right, so I'm assuming you've at least given it a shot. So let's work through it together. So the key here is to realize, well okay, this point, this base right over here, this point right over here, it's halfway in this direction, and halfway in this direction. So we can figure out, well this entire, this entire length right over here, is length four, so halfway, this is going to be, I'll write it with perspective, that's going to be two, and that's going to be two, just like that. And then the other thing we can figure out, we can figure out what this length is going to be, because once again, it's halfway in that direction.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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So let's work through it together. So the key here is to realize, well okay, this point, this base right over here, this point right over here, it's halfway in this direction, and halfway in this direction. So we can figure out, well this entire, this entire length right over here, is length four, so halfway, this is going to be, I'll write it with perspective, that's going to be two, and that's going to be two, just like that. And then the other thing we can figure out, we can figure out what this length is going to be, because once again, it's halfway in that direction. So if this whole thing is two, and we see it right over here, this is a rectangular prism, so this length is going to be the same thing as this length. So if this whole thing is two, then each of these, this is going to be one, and this is going to be one right over there. Well how does that help us?
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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And then the other thing we can figure out, we can figure out what this length is going to be, because once again, it's halfway in that direction. So if this whole thing is two, and we see it right over here, this is a rectangular prism, so this length is going to be the same thing as this length. So if this whole thing is two, then each of these, this is going to be one, and this is going to be one right over there. Well how does that help us? Well using that information, we should be able to figure out, we should be able to figure out this length. Actually let me, actually I'll keep it in this color, because this color is easy to see. We should be able to figure out this length.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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Well how does that help us? Well using that information, we should be able to figure out, we should be able to figure out this length. Actually let me, actually I'll keep it in this color, because this color is easy to see. We should be able to figure out this length. Well why is this length interesting? Well if we know that length, that length forms a right triangle. That length and the one are the two non-hypotenuse sides of a right triangle, and then the x would be the hypotenuse.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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We should be able to figure out this length. Well why is this length interesting? Well if we know that length, that length forms a right triangle. That length and the one are the two non-hypotenuse sides of a right triangle, and then the x would be the hypotenuse. So we could just apply the Pythagorean theorem. So if we can figure out this, we can figure out x. So let's do one step at a time.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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That length and the one are the two non-hypotenuse sides of a right triangle, and then the x would be the hypotenuse. So we could just apply the Pythagorean theorem. So if we can figure out this, we can figure out x. So let's do one step at a time. How do we figure out, how do we figure out, I don't know, let me call this length, how do we call that, how do we figure out length a? Well let's just take it out, and look at it in two dimensions. So if we look at it in two dimensions, if we look at it in two dimensions, it would look something like this.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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So let's do one step at a time. How do we figure out, how do we figure out, I don't know, let me call this length, how do we call that, how do we figure out length a? Well let's just take it out, and look at it in two dimensions. So if we look at it in two dimensions, if we look at it in two dimensions, it would look something like this. So that's our length a. We know that this length is half of this side right over here, so that's going to be one. And actually let me do it with the same colors.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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So if we look at it in two dimensions, if we look at it in two dimensions, it would look something like this. So that's our length a. We know that this length is half of this side right over here, so that's going to be one. And actually let me do it with the same colors. So this right over here is the same thing as this right over here, and it is going to be of length one. And then this right over here is going to be the same as this right over here, which is going to be of length two. And so we can just use the Pythagorean theorem here.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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And actually let me do it with the same colors. So this right over here is the same thing as this right over here, and it is going to be of length one. And then this right over here is going to be the same as this right over here, which is going to be of length two. And so we can just use the Pythagorean theorem here. We know that the hypotenuse squared is going to be equal to one squared, one squared plus two squared. One squared plus two squared, which of course is equal to five, sorry, is equal to this one plus four, which is equal to five. So we could write a, let me do this in the magenta color, we can write a squared is equal to five, or we could say that a is equal to the principal root of five.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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And so we can just use the Pythagorean theorem here. We know that the hypotenuse squared is going to be equal to one squared, one squared plus two squared. One squared plus two squared, which of course is equal to five, sorry, is equal to this one plus four, which is equal to five. So we could write a, let me do this in the magenta color, we can write a squared is equal to five, or we could say that a is equal to the principal root of five. So this length right over here is the square root of five, the principal root of five. And now we can use that information to solve for x. And let's take this right triangle, and it takes a little bit of visualization practice to visualize this, right?
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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So we could write a, let me do this in the magenta color, we can write a squared is equal to five, or we could say that a is equal to the principal root of five. So this length right over here is the square root of five, the principal root of five. And now we can use that information to solve for x. And let's take this right triangle, and it takes a little bit of visualization practice to visualize this, right? But notice this is a right triangle. This height right here of length one is perpendicular to this entire plane. So let's see if I can draw it.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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And let's take this right triangle, and it takes a little bit of visualization practice to visualize this, right? But notice this is a right triangle. This height right here of length one is perpendicular to this entire plane. So let's see if I can draw it. So this, so we have this side is the square root of five, and then we have a height. Let me do it, it looks like it's in a, well, let's say we're talking about this height right over here. This height right over here is length one.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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So let's see if I can draw it. So this, so we have this side is the square root of five, and then we have a height. Let me do it, it looks like it's in a, well, let's say we're talking about this height right over here. This height right over here is length one. So I'll draw that. That is of length one. And we are trying to figure out x.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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This height right over here is length one. So I'll draw that. That is of length one. And we are trying to figure out x. We are trying to figure out x in orange. So we're trying to figure out, we're trying to figure out this x. And once again, we know this is a right triangle.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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And we are trying to figure out x. We are trying to figure out x in orange. So we're trying to figure out, we're trying to figure out this x. And once again, we know this is a right triangle. This is a right triangle, so we can apply the Pythagorean Theorem again. So we will have x squared is equal to one squared, which is just, I'll just write it one squared plus square root of five squared, plus square root of five squared. Well, that gets us x squared is equal to one plus five, right, square root of five squared is just five.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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And once again, we know this is a right triangle. This is a right triangle, so we can apply the Pythagorean Theorem again. So we will have x squared is equal to one squared, which is just, I'll just write it one squared plus square root of five squared, plus square root of five squared. Well, that gets us x squared is equal to one plus five, right, square root of five squared is just five. So one plus five is equal to six. So we get x is equal to the square root of six. And we're done.
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Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3
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We're gonna move, it's kinda small, I hope you can see it on your video screen. We're gonna move positive eight. Every point here is gonna move positive eight in the x direction. It's x coordinate is going to increase by eight, and, or the corresponding point in the image is x coordinate is going to increase by eight, and the corresponding point in the image's y coordinate is going to decrease by one. So let's do that. And I'll focus on the vertices. Whoops, let me drag that to the trash.
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Drawing image of translation Transformations Geometry Khan Academy.mp3
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It's x coordinate is going to increase by eight, and, or the corresponding point in the image is x coordinate is going to increase by eight, and the corresponding point in the image's y coordinate is going to decrease by one. So let's do that. And I'll focus on the vertices. Whoops, let me drag that to the trash. I didn't mean to do that. I'm gonna focus on the vertices, because that's, well, that's the easiest thing for my brain to work with. So, and actually this is what the tool expects as well.
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Drawing image of translation Transformations Geometry Khan Academy.mp3
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Whoops, let me drag that to the trash. I didn't mean to do that. I'm gonna focus on the vertices, because that's, well, that's the easiest thing for my brain to work with. So, and actually this is what the tool expects as well. So the point B is gonna move eight to the right, or its corresponding point in the image is gonna have an x coordinate eight larger. So right now the x coordinate is negative four. If you added eight to that, it would be positive four.
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Drawing image of translation Transformations Geometry Khan Academy.mp3
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So, and actually this is what the tool expects as well. So the point B is gonna move eight to the right, or its corresponding point in the image is gonna have an x coordinate eight larger. So right now the x coordinate is negative four. If you added eight to that, it would be positive four. And its y coordinate is gonna be one lower. Right now point B's y coordinate is eight. One lower than that is seven.
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Drawing image of translation Transformations Geometry Khan Academy.mp3
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If you added eight to that, it would be positive four. And its y coordinate is gonna be one lower. Right now point B's y coordinate is eight. One lower than that is seven. So in the image, the corresponding point in the image would be right over there. And you see we moved eight to the right and one down. Let's do that with point C. It's at x equals negative seven.
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Drawing image of translation Transformations Geometry Khan Academy.mp3
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One lower than that is seven. So in the image, the corresponding point in the image would be right over there. And you see we moved eight to the right and one down. Let's do that with point C. It's at x equals negative seven. If you move eight to the right, if you increase your x coordinate by eight, you're gonna move to x equals one. And then if you change your y coordinate by negative one, you're gonna move down one, and you're gonna get to that point right over there. Now let's do it with point A.
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Drawing image of translation Transformations Geometry Khan Academy.mp3
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Let's do that with point C. It's at x equals negative seven. If you move eight to the right, if you increase your x coordinate by eight, you're gonna move to x equals one. And then if you change your y coordinate by negative one, you're gonna move down one, and you're gonna get to that point right over there. Now let's do it with point A. So point A's x coordinate is negative one. If you add eight to it, it's going to be positive seven. And its current y coordinate is two.
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Drawing image of translation Transformations Geometry Khan Academy.mp3
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Now let's do it with point A. So point A's x coordinate is negative one. If you add eight to it, it's going to be positive seven. And its current y coordinate is two. If you take one away from it, you're gonna get to a y coordinate of one. And so there you have it. Let's see, how do I connect these two?
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Drawing image of translation Transformations Geometry Khan Academy.mp3
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And its current y coordinate is two. If you take one away from it, you're gonna get to a y coordinate of one. And so there you have it. Let's see, how do I connect these two? Oh, there you go. And we can check our answer. And we got it right.
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Drawing image of translation Transformations Geometry Khan Academy.mp3
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So pause this video and see if you can work this out on your own before we work through this together. All right, now let's work through this together, and it looks like for every one of these, or actually almost every one of these, they've given us two angles and they've given us a side. This triangle IJH, they've only given us two angles. So what I'd like to do is if I know two angles of a triangle, I can figure out the third angle because the sum of the angles of a triangle have to add up to 180 degrees. And then I can use that information, maybe with the sides that they give us, in order to judge which of these triangles are congruent. So first of all, what is going to be the measure of this angle right over here, the measure of angle ACB? Pause the video and try to think about that.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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So what I'd like to do is if I know two angles of a triangle, I can figure out the third angle because the sum of the angles of a triangle have to add up to 180 degrees. And then I can use that information, maybe with the sides that they give us, in order to judge which of these triangles are congruent. So first of all, what is going to be the measure of this angle right over here, the measure of angle ACB? Pause the video and try to think about that. Well, one way to think about it, if we call the measure of that angle X, we know that X plus 36 plus 82 needs to be equal to 180. I'm just giving their measures in degrees here. And so you could say X plus, let's see, 36 plus 82 is 118.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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Pause the video and try to think about that. Well, one way to think about it, if we call the measure of that angle X, we know that X plus 36 plus 82 needs to be equal to 180. I'm just giving their measures in degrees here. And so you could say X plus, let's see, 36 plus 82 is 118. Did I do that right? Six plus two is eight, and then three plus eight is 11. Yep, that's right.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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And so you could say X plus, let's see, 36 plus 82 is 118. Did I do that right? Six plus two is eight, and then three plus eight is 11. Yep, that's right. So that's going to be equal to 180. And then if I subtract 118 from both sides, I'm going to get X is equal to, 180 minus 18 is 62. So this is X is equal to 62, or this is a 62-degree angle, I guess is another way of thinking about it.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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Yep, that's right. So that's going to be equal to 180. And then if I subtract 118 from both sides, I'm going to get X is equal to, 180 minus 18 is 62. So this is X is equal to 62, or this is a 62-degree angle, I guess is another way of thinking about it. I could put everything in terms of degrees if you like. All right, now let's do the same thing with this one right over here. Well, this one has an 82-degree angle and a 62-degree angle, just like this triangle over here.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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So this is X is equal to 62, or this is a 62-degree angle, I guess is another way of thinking about it. I could put everything in terms of degrees if you like. All right, now let's do the same thing with this one right over here. Well, this one has an 82-degree angle and a 62-degree angle, just like this triangle over here. So we know that the third angle needs to be 36 degrees. 36 degrees, because we know 82 and 62, if you need to get to 180, it has to be 36. We just figured that out from this first triangle over here.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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Well, this one has an 82-degree angle and a 62-degree angle, just like this triangle over here. So we know that the third angle needs to be 36 degrees. 36 degrees, because we know 82 and 62, if you need to get to 180, it has to be 36. We just figured that out from this first triangle over here. Now if we look over here, 36 degrees and 59, this definitely looks like it has different angles, but let's figure out what this angle would have to be. So if we call that Y degrees, we know, I'll do it over here, Y plus 36 plus 59 is equal to 180. And I'm just thinking in terms of degrees here.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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We just figured that out from this first triangle over here. Now if we look over here, 36 degrees and 59, this definitely looks like it has different angles, but let's figure out what this angle would have to be. So if we call that Y degrees, we know, I'll do it over here, Y plus 36 plus 59 is equal to 180. And I'm just thinking in terms of degrees here. So Y plus, this is going to be equal to, what is this? This is going to be equal to 95 is equal to 180. Did I do that right?
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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And I'm just thinking in terms of degrees here. So Y plus, this is going to be equal to, what is this? This is going to be equal to 95 is equal to 180. Did I do that right? Yep, that's 80 plus 15, yep, 95. And then if I subtract 95 from both sides, what am I left with? I'm left with Y is equal to 85 degrees.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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