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And then you have this like this. So right off the bat, it's definitely a quadrilateral. I have four sides. But the key question, are any of these sides parallel to any of the other sides? So just looking at it, side CB is clearly not parallel to AD. You just can look at it. Now, it also looks like CD is not parallel to BA.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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But the key question, are any of these sides parallel to any of the other sides? So just looking at it, side CB is clearly not parallel to AD. You just can look at it. Now, it also looks like CD is not parallel to BA. But maybe I just drew it badly. Maybe they actually are parallel. So let's see if we can verify that.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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Now, it also looks like CD is not parallel to BA. But maybe I just drew it badly. Maybe they actually are parallel. So let's see if we can verify that. So the way to tell whether two things are parallel is to actually figure out their slopes. So let's first figure out the slope. Let's figure out our slope of AB or BA.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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So let's see if we can verify that. So the way to tell whether two things are parallel is to actually figure out their slopes. So let's first figure out the slope. Let's figure out our slope of AB or BA. So let's figure out our slope here. Your slope is going to be the change in y over your change in x. And in this case, you could think of it as we're starting at the point negative 5, 2, and we're ending at the point 1, 6.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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Let's figure out our slope of AB or BA. So let's figure out our slope here. Your slope is going to be the change in y over your change in x. And in this case, you could think of it as we're starting at the point negative 5, 2, and we're ending at the point 1, 6. So what's our change in y? We go from 2 all the way to 6. Or you could say it's 6 minus 2.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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And in this case, you could think of it as we're starting at the point negative 5, 2, and we're ending at the point 1, 6. So what's our change in y? We go from 2 all the way to 6. Or you could say it's 6 minus 2. Our change in y is 4. What's our change in x? Well, we go from negative 5 to 1.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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Or you could say it's 6 minus 2. Our change in y is 4. What's our change in x? Well, we go from negative 5 to 1. So we increased by 6. Or another way of thinking about it, 1 minus negative 5. Well, that's going to be equal to 6.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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Well, we go from negative 5 to 1. So we increased by 6. Or another way of thinking about it, 1 minus negative 5. Well, that's going to be equal to 6. So our slope here is 2 thirds. It's 2 over 3. Every time we move 3 in the x direction, we go up 2 in the y direction.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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Well, that's going to be equal to 6. So our slope here is 2 thirds. It's 2 over 3. Every time we move 3 in the x direction, we go up 2 in the y direction. Move 3 in the x direction, go up 2 in the y direction. Now let's think about line CD up here. What is our slope?
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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Every time we move 3 in the x direction, we go up 2 in the y direction. Move 3 in the x direction, go up 2 in the y direction. Now let's think about line CD up here. What is our slope? Our change in y over change in x is going to be equal to, let's see, our change in, let's figure out our change in x first. Our change in x, we're going from negative 7, 8 all the way to 2, 11. So our change in x, we're going from negative 7 to 2.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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What is our slope? Our change in y over change in x is going to be equal to, let's see, our change in, let's figure out our change in x first. Our change in x, we're going from negative 7, 8 all the way to 2, 11. So our change in x, we're going from negative 7 to 2. Or we could say 2 minus negative 7. So we are increasing by 9. That's our change in x.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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So our change in x, we're going from negative 7 to 2. Or we could say 2 minus negative 7. So we are increasing by 9. That's our change in x. So that's going to be equal to 9. And then our change in y, well, it looks like we've gone up. We've gone from 8 to 11.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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That's our change in x. So that's going to be equal to 9. And then our change in y, well, it looks like we've gone up. We've gone from 8 to 11. So we've gone up 3. Or we could say 11 minus 8. Notice, end point minus start point.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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We've gone from 8 to 11. So we've gone up 3. Or we could say 11 minus 8. Notice, end point minus start point. You have to do that on both the top and the bottom. Otherwise, you're not going to actually be calculating your change in y over change in x. But you notice our change, when our x increases by 9, our y increased by 3.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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Notice, end point minus start point. You have to do that on both the top and the bottom. Otherwise, you're not going to actually be calculating your change in y over change in x. But you notice our change, when our x increases by 9, our y increased by 3. So the slope here is equal to 1 third. So these actually have different slopes. So none of these lines are parallel to each other.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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But you notice our change, when our x increases by 9, our y increased by 3. So the slope here is equal to 1 third. So these actually have different slopes. So none of these lines are parallel to each other. So this isn't even a parallelogram. This isn't even a trapezoid. Parallelogram, you have to have two pairs of parallel sides.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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So none of these lines are parallel to each other. So this isn't even a parallelogram. This isn't even a trapezoid. Parallelogram, you have to have two pairs of parallel sides. Trapezoid, you have to have one pair of parallel sides. This isn't the case for any of these. Or none of these sides are parallel.
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Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3
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How would we do that? Well, the way that we can approach it is by creating what will eventually be a transversal between the two parallel lines. So let me draw that. So I'm just drawing a line that goes through my point and intersects my original line. I'm doing that, so it's going to look like that. And then I'm really just going to use the idea of corresponding angle congruence for parallel lines. So what I can do is now take my compass and think about this angle right over here.
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Geometric constructions parallel line Congruence High school geometry Khan Academy.mp3
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So I'm just drawing a line that goes through my point and intersects my original line. I'm doing that, so it's going to look like that. And then I'm really just going to use the idea of corresponding angle congruence for parallel lines. So what I can do is now take my compass and think about this angle right over here. So I'll draw it like that and say, all right, if I draw an arc of the same radius over here, can I reconstruct that angle? And so where should the point be on this left end? Well, to do that, I can just measure the distance between these two points using my compass.
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Geometric constructions parallel line Congruence High school geometry Khan Academy.mp3
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So what I can do is now take my compass and think about this angle right over here. So I'll draw it like that and say, all right, if I draw an arc of the same radius over here, can I reconstruct that angle? And so where should the point be on this left end? Well, to do that, I can just measure the distance between these two points using my compass. So I'm adjusting it a little bit to get the distance between those two points. And then I can use that up over here to figure out, I got a little bit shaky, I can figure out that point right over there. And just like that, I now have two corresponding angles defined by a transversal and parallel lines.
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Geometric constructions parallel line Congruence High school geometry Khan Academy.mp3
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Well, to do that, I can just measure the distance between these two points using my compass. So I'm adjusting it a little bit to get the distance between those two points. And then I can use that up over here to figure out, I got a little bit shaky, I can figure out that point right over there. And just like that, I now have two corresponding angles defined by a transversal and parallel lines. So what I could do is take my straight edge and make it go through those points that I just created. So let's see, make sure I'm going through them. And it would look like that.
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Geometric constructions parallel line Congruence High school geometry Khan Academy.mp3
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And just like that, I now have two corresponding angles defined by a transversal and parallel lines. So what I could do is take my straight edge and make it go through those points that I just created. So let's see, make sure I'm going through them. And it would look like that. And I have just constructed two parallel lines. And once again, how do I know that this line is parallel to this line? Because we have a transversal that intersects both of them.
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Geometric constructions parallel line Congruence High school geometry Khan Academy.mp3
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we're asked to convert 150 degrees and negative 45 degrees to radians. So let's think about the relationship between degrees and radians. And to do that, let me just draw a little circle here. So that's the center of the circle. And then do my best shot, best attempt to freehand draw a reasonable looking circle. Yeah, that's not, I've done worse. I've done worse than that.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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So that's the center of the circle. And then do my best shot, best attempt to freehand draw a reasonable looking circle. Yeah, that's not, I've done worse. I've done worse than that. All right. Now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that? Well, we know that that would be 360 degrees.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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I've done worse than that. All right. Now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that? Well, we know that that would be 360 degrees. Well, if we did the same thing, how many radians is that? If we were to go all the way around the circle? Well, we just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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Well, we know that that would be 360 degrees. Well, if we did the same thing, how many radians is that? If we were to go all the way around the circle? Well, we just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle. So if you go all the way around, you're really talking about the arc length of the entire circle, or essentially the circumference of the circle. And you're essentially saying how many radiuses this is, or radii, or how many radii is the circumference of the circle? You know a circumference of a circle is two pi, is two pi times the radius.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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Well, we just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle. So if you go all the way around, you're really talking about the arc length of the entire circle, or essentially the circumference of the circle. And you're essentially saying how many radiuses this is, or radii, or how many radii is the circumference of the circle? You know a circumference of a circle is two pi, is two pi times the radius. Or you could say that the length of the circumference of the circle is two pi radii. Two pi radii. If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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You know a circumference of a circle is two pi, is two pi times the radius. Or you could say that the length of the circumference of the circle is two pi radii. Two pi radii. If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi. That just comes from, really actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle. So if we were to go all the way around this, this is also two pi radians. This is also two pi, two pi radians.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi. That just comes from, really actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle. So if we were to go all the way around this, this is also two pi radians. This is also two pi, two pi radians. So that tells us that two pi radians, as an angle measure, is the exact same thing, and I'm gonna write it out as 360, 360 degrees. And then we can take all of this relationship and manipulate it in different ways. If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case you are left with, if you divide both sides by two, you are left with pi radians, pi radians is equal to 180 degrees, is equal to 180 degrees, 180 degrees.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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This is also two pi, two pi radians. So that tells us that two pi radians, as an angle measure, is the exact same thing, and I'm gonna write it out as 360, 360 degrees. And then we can take all of this relationship and manipulate it in different ways. If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case you are left with, if you divide both sides by two, you are left with pi radians, pi radians is equal to 180 degrees, is equal to 180 degrees, 180 degrees. So how can we use this relationship now to figure out what 150 degrees is? Well, this relationship we could write in different ways. We could divide both sides by 180 degrees, and we could get pi radians, pi radians over 180 degrees, over 180 degrees is equal to one, which is just another way of saying that there are pi radians for every 180 degrees, or you could say pi over 180 radians per degree.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case you are left with, if you divide both sides by two, you are left with pi radians, pi radians is equal to 180 degrees, is equal to 180 degrees, 180 degrees. So how can we use this relationship now to figure out what 150 degrees is? Well, this relationship we could write in different ways. We could divide both sides by 180 degrees, and we could get pi radians, pi radians over 180 degrees, over 180 degrees is equal to one, which is just another way of saying that there are pi radians for every 180 degrees, or you could say pi over 180 radians per degree. The other option, you could divide both sides of this by pi radians. You could say, you would get, on the left-hand side, you get one, and you would also get, on the right-hand side, you would get 180 degrees for every pi radians, 180 degrees for every pi radians, for every pi radians, or you could interpret this as 180 over pi degrees per radian. So how would we figure out, how would we do what they asked us?
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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We could divide both sides by 180 degrees, and we could get pi radians, pi radians over 180 degrees, over 180 degrees is equal to one, which is just another way of saying that there are pi radians for every 180 degrees, or you could say pi over 180 radians per degree. The other option, you could divide both sides of this by pi radians. You could say, you would get, on the left-hand side, you get one, and you would also get, on the right-hand side, you would get 180 degrees for every pi radians, 180 degrees for every pi radians, for every pi radians, or you could interpret this as 180 over pi degrees per radian. So how would we figure out, how would we do what they asked us? So let's convert 150 degrees to radians. Let me write the word out. So 150 degrees, 150 degrees.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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So how would we figure out, how would we do what they asked us? So let's convert 150 degrees to radians. Let me write the word out. So 150 degrees, 150 degrees. Well, we wanna convert this to radians, so we really care about how many radians there are, how many radians there are per degree. Actually, let me do that in that color. So we care about how many radians, radians there are per degree, per, let me do that same green color, per degree.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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So 150 degrees, 150 degrees. Well, we wanna convert this to radians, so we really care about how many radians there are, how many radians there are per degree. Actually, let me do that in that color. So we care about how many radians, radians there are per degree, per, let me do that same green color, per degree. Well, how many radians are there per degree? We already know, there's pi radians for every 180 degrees, or there are pi, pi, let me do that yellow color, there are pi over 180, pi over 180 radians per degree. And so if we multiply, and this all works out, because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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So we care about how many radians, radians there are per degree, per, let me do that same green color, per degree. Well, how many radians are there per degree? We already know, there's pi radians for every 180 degrees, or there are pi, pi, let me do that yellow color, there are pi over 180, pi over 180 radians per degree. And so if we multiply, and this all works out, because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians. So what do we get? Well, this becomes, let me just rewrite it, 150 times pi, times pi, all of that over 180, all of that over 180, so this is equal to, and we get it in radians, radians. And so if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like 30.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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And so if we multiply, and this all works out, because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians. So what do we get? Well, this becomes, let me just rewrite it, 150 times pi, times pi, all of that over 180, all of that over 180, so this is equal to, and we get it in radians, radians. And so if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like 30. So if you divide the numerator by 30, you get five. You divide the denominator by 30, you get six. So you get five pi over six radians, or five six pi radians, depending how you want to do it.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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And so if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like 30. So if you divide the numerator by 30, you get five. You divide the denominator by 30, you get six. So you get five pi over six radians, or five six pi radians, depending how you want to do it. Now let's do the same thing for negative 45 degrees. What do you get for negative 45 degrees if you were to convert that to radians? Well, same exact process.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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So you get five pi over six radians, or five six pi radians, depending how you want to do it. Now let's do the same thing for negative 45 degrees. What do you get for negative 45 degrees if you were to convert that to radians? Well, same exact process. You have negative, and I'll do this one a little quicker, for negative 45 degrees, I'll write down the word, times pi radians, pi radians for every 180 degrees, for every 180 degrees, the degrees cancel out, and you are left with negative 45 pi over 180 radians. So this is equal to negative 45 pi, pi over 180, over 180 radians, radians. So how can we simplify this?
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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Well, same exact process. You have negative, and I'll do this one a little quicker, for negative 45 degrees, I'll write down the word, times pi radians, pi radians for every 180 degrees, for every 180 degrees, the degrees cancel out, and you are left with negative 45 pi over 180 radians. So this is equal to negative 45 pi, pi over 180, over 180 radians, radians. So how can we simplify this? Well, it looks like they're both, at minimum, divisible by nine. Nine times five is 45, this is nine times 20, so actually it's gonna be divisible by more than just, let's see, nine, nine, they're both, well, I will just, actually, they're both divisible by 45, what am I doing? Okay, so if you divide the numerator by 45, you get one.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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So how can we simplify this? Well, it looks like they're both, at minimum, divisible by nine. Nine times five is 45, this is nine times 20, so actually it's gonna be divisible by more than just, let's see, nine, nine, they're both, well, I will just, actually, they're both divisible by 45, what am I doing? Okay, so if you divide the numerator by 45, you get one. You divide the denominator by 45, 45 goes into 180 four times, four times. You're left with negative pi over four radians. This is equal to negative pi over four radians, and we are done.
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Example Converting degrees to radians Trigonometry Khan Academy.mp3
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And I encourage you to pause this video and try to think about it on your own. So let's think about what the change in x that this translation does and the change in y that this translation does. So our change in x, our change in x is going to be our ending point, our ending point. So negative 203 minus our starting point minus negative 169, minus negative 169. And this is the same thing as, I'll just write it in yellow, negative 203 plus 169, which is the same thing as 169 minus 203. I'm just writing it this way for my brain to process it, which is the same thing as the negative of 203 minus 169. And let's see, this one is the easiest for my brain to process.
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Determining a translation between points Transformations Geometry Khan Academy.mp3
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So negative 203 minus our starting point minus negative 169, minus negative 169. And this is the same thing as, I'll just write it in yellow, negative 203 plus 169, which is the same thing as 169 minus 203. I'm just writing it this way for my brain to process it, which is the same thing as the negative of 203 minus 169. And let's see, this one is the easiest for my brain to process. This is going to be 34, so this is going to be negative 34. That's our change in x. Now let's think about our change in y.
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Determining a translation between points Transformations Geometry Khan Academy.mp3
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And let's see, this one is the easiest for my brain to process. This is going to be 34, so this is going to be negative 34. That's our change in x. Now let's think about our change in y. Our change in y is, our ending point is negative 68, negative 68, and our starting point is 434. So our ending point minus our starting point, minus 434, is going to be equal to, so this is just going to be the same thing as the negative, of 68 plus 434. And let's see, 430 plus 60 is 490 plus another eight plus four, plus another 12, so that's going to get us to 502.
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Determining a translation between points Transformations Geometry Khan Academy.mp3
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Now let's think about our change in y. Our change in y is, our ending point is negative 68, negative 68, and our starting point is 434. So our ending point minus our starting point, minus 434, is going to be equal to, so this is just going to be the same thing as the negative, of 68 plus 434. And let's see, 430 plus 60 is 490 plus another eight plus four, plus another 12, so that's going to get us to 502. This is going to be negative 502. So this translation, what it does is, it shifts us in the x direction by negative 34, and it shifts us in the y direction by negative 502. So if we're starting at 31 comma negative 529, we're going to end up at, let me write it this way.
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Determining a translation between points Transformations Geometry Khan Academy.mp3
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And let's see, 430 plus 60 is 490 plus another eight plus four, plus another 12, so that's going to get us to 502. This is going to be negative 502. So this translation, what it does is, it shifts us in the x direction by negative 34, and it shifts us in the y direction by negative 502. So if we're starting at 31 comma negative 529, we're going to end up at, let me write it this way. So if we're starting at this point, if we're starting at 31 comma negative 529, we're going to end up at 31 minus 34, minus 34, that's how much we're going to shift it, and negative 529, negative 529, so negative 529 minus 502, minus 502. So what's this going to be? 31 minus 34 is negative three, and then if we have negative 529 minus 502, that's going to be negative 1031.
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Determining a translation between points Transformations Geometry Khan Academy.mp3
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For each of these diagrams, I want to think about whether this blue line represents an axis of symmetry. And the way we can tell is if on both sides of the blue line, we essentially have mirror images. So let's imagine, let's take this top part of this polygon, the part that is above this blue line here. And let's reflect it across the blue line. You could almost imagine that it's a reflection over some type of a lake or something and see if we get exactly what we have below. Then this would be an axis of symmetry. So this point right over here, this distance to the blue line, let's go the same amount on the other side would get you right there.
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Axis of symmetry Transformations Geometry Khan Academy.mp3
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And let's reflect it across the blue line. You could almost imagine that it's a reflection over some type of a lake or something and see if we get exactly what we have below. Then this would be an axis of symmetry. So this point right over here, this distance to the blue line, let's go the same amount on the other side would get you right there. And so you immediately see, we start ending up with a point that is off what's actually here in black, the actual bottom part of the polygon. So this is a pretty good clue that this is not an axis of symmetry. But let's just continue it, just to go through the exercise.
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Axis of symmetry Transformations Geometry Khan Academy.mp3
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So this point right over here, this distance to the blue line, let's go the same amount on the other side would get you right there. And so you immediately see, we start ending up with a point that is off what's actually here in black, the actual bottom part of the polygon. So this is a pretty good clue that this is not an axis of symmetry. But let's just continue it, just to go through the exercise. So this point, if you reflected it across this blue line, would get you here. This point, we'll do it in a different color. This point, if you were to reflect it across this blue line, it would get you.
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Axis of symmetry Transformations Geometry Khan Academy.mp3
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But let's just continue it, just to go through the exercise. So this point, if you reflected it across this blue line, would get you here. This point, we'll do it in a different color. This point, if you were to reflect it across this blue line, it would get you. Make sure I can do that relatively straight. I can do a straighter job than that. So if you go about that distance about it, and I want to go straight down into the blue line, and say I'm going to go the same distance on the other side, it gets me to right around there.
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Axis of symmetry Transformations Geometry Khan Academy.mp3
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This point, if you were to reflect it across this blue line, it would get you. Make sure I can do that relatively straight. I can do a straighter job than that. So if you go about that distance about it, and I want to go straight down into the blue line, and say I'm going to go the same distance on the other side, it gets me to right around there. And then this point over here, if I were to drop it straight down, then if I were to go the same distance on the other side, it gets me right around there. And then finally, this point gets me right around there. So its mirror image of this top part would look something like this.
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Axis of symmetry Transformations Geometry Khan Academy.mp3
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So if you go about that distance about it, and I want to go straight down into the blue line, and say I'm going to go the same distance on the other side, it gets me to right around there. And then this point over here, if I were to drop it straight down, then if I were to go the same distance on the other side, it gets me right around there. And then finally, this point gets me right around there. So its mirror image of this top part would look something like this. My best attempt to draw it would look something like this, which is very different than the part of the polygon that's actually on the other side of this blue line. So in this case, the blue line is not an axis of symmetry. So this is no, no.
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Axis of symmetry Transformations Geometry Khan Academy.mp3
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So its mirror image of this top part would look something like this. My best attempt to draw it would look something like this, which is very different than the part of the polygon that's actually on the other side of this blue line. So in this case, the blue line is not an axis of symmetry. So this is no, no. This blue line is not an axis of symmetry. Now let's look at it over here. And our eyes pick this out very naturally.
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Axis of symmetry Transformations Geometry Khan Academy.mp3
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So this is no, no. This blue line is not an axis of symmetry. Now let's look at it over here. And our eyes pick this out very naturally. Here you can see that it looks like this blue line really divides this polygon in half. It really does look like mirror images. It really does look like, if you imagine that this is some type of a lake, a still lake, so I shouldn't actually draw waves, but this is some type of a lake, that this is the reflection.
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Axis of symmetry Transformations Geometry Khan Academy.mp3
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And our eyes pick this out very naturally. Here you can see that it looks like this blue line really divides this polygon in half. It really does look like mirror images. It really does look like, if you imagine that this is some type of a lake, a still lake, so I shouldn't actually draw waves, but this is some type of a lake, that this is the reflection. And we can even go point by point here. So this point right over here is the same distance from if we dropped a perpendicular to this point as this one right over here. This one over here, same distance as this point right over here.
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Axis of symmetry Transformations Geometry Khan Academy.mp3
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Well, we're going to prove in this video, it's a couple of fairly straightforward parallelogram related proofs. And this first one we're gonna say, hey, if we have this parallelogram ABCD, let's prove that the opposite sides have the same length. So prove that AB is equal to DC, and that AD is equal to BC. So let me draw a diagonal here. So I'm gonna draw a diagonal. And this diagonal, depending on how you view it, is intersecting two sets of parallel lines. So you could also consider it to be a transversal.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So let me draw a diagonal here. So I'm gonna draw a diagonal. And this diagonal, depending on how you view it, is intersecting two sets of parallel lines. So you could also consider it to be a transversal. Actually, let me draw it a little bit neater than that. I can do a better job. So, nope, that's not any better.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So you could also consider it to be a transversal. Actually, let me draw it a little bit neater than that. I can do a better job. So, nope, that's not any better. That is about as good as I can do. So if we view DB, this diagonal DB, we can view it as a transversal for the parallel lines AB and DC. And if you view it that way, you can pick out that angle ABD is going to be congruent.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So, nope, that's not any better. That is about as good as I can do. So if we view DB, this diagonal DB, we can view it as a transversal for the parallel lines AB and DC. And if you view it that way, you can pick out that angle ABD is going to be congruent. So angle ABD, that's that angle right there, is going to be congruent to angle BDC because they're alternate interior angles. You have a transversal, parallel lines. So we know that angle ABD is going to be congruent to angle BDC.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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And if you view it that way, you can pick out that angle ABD is going to be congruent. So angle ABD, that's that angle right there, is going to be congruent to angle BDC because they're alternate interior angles. You have a transversal, parallel lines. So we know that angle ABD is going to be congruent to angle BDC. Angle BDC. Now, you could also view this diagonal DB, you could view it as a transversal of these two parallel lines, of the other pair of parallel lines, AD and BC. And if you look at it that way, then you immediately see that angle DBC, angle DBC right over here, angle DBC is going to be congruent to angle ADB, to angle ADB, angle ADB, for the exact same reason.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So we know that angle ABD is going to be congruent to angle BDC. Angle BDC. Now, you could also view this diagonal DB, you could view it as a transversal of these two parallel lines, of the other pair of parallel lines, AD and BC. And if you look at it that way, then you immediately see that angle DBC, angle DBC right over here, angle DBC is going to be congruent to angle ADB, to angle ADB, angle ADB, for the exact same reason. They are alternate interior angles of a transversal intersecting these two parallel lines. So I could write this. This is alternate interior angles, interior angles are congruent when you have a transversal intersecting two parallel lines.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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And if you look at it that way, then you immediately see that angle DBC, angle DBC right over here, angle DBC is going to be congruent to angle ADB, to angle ADB, angle ADB, for the exact same reason. They are alternate interior angles of a transversal intersecting these two parallel lines. So I could write this. This is alternate interior angles, interior angles are congruent when you have a transversal intersecting two parallel lines. And we also see that both of these triangles, triangle ADB and triangle CDB, both share this side over here. It's obviously equal to itself. Now, why is this useful?
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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This is alternate interior angles, interior angles are congruent when you have a transversal intersecting two parallel lines. And we also see that both of these triangles, triangle ADB and triangle CDB, both share this side over here. It's obviously equal to itself. Now, why is this useful? Well, you might realize that we've just shown that both of these triangles, they have this pink angle, then they have this side in common, and then they have the green angle. Pink angle, side in common, and then the green angle. So we've just shown by angle, side, angle, that these two triangles are congruent.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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Now, why is this useful? Well, you might realize that we've just shown that both of these triangles, they have this pink angle, then they have this side in common, and then they have the green angle. Pink angle, side in common, and then the green angle. So we've just shown by angle, side, angle, that these two triangles are congruent. So let me write this down. We have shown that triangle, I'll go from non-labeled to pink to green. ADB is congruent to triangle non-labeled to pink to green.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So we've just shown by angle, side, angle, that these two triangles are congruent. So let me write this down. We have shown that triangle, I'll go from non-labeled to pink to green. ADB is congruent to triangle non-labeled to pink to green. Non-labeled to pink to green. C, CBD, CBD. And this comes out of angle, side, angle congruency.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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ADB is congruent to triangle non-labeled to pink to green. Non-labeled to pink to green. C, CBD, CBD. And this comes out of angle, side, angle congruency. So this is from angle, side, angle, angle, side, angle, congruency. Well, what does that do for us? Well, if two triangles are congruent, then all of the corresponding features of the two triangles are going to be congruent.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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And this comes out of angle, side, angle congruency. So this is from angle, side, angle, angle, side, angle, congruency. Well, what does that do for us? Well, if two triangles are congruent, then all of the corresponding features of the two triangles are going to be congruent. In particular, side DC, side DC corresponds to side BA, side DC on this bottom triangle corresponds to side BA on that top triangle. So they need to be congruent. So DC, so we get DC is going to be equal to BA, and that's because they are corresponding sides, corresponding sides of congruent, congruent triangles.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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Well, if two triangles are congruent, then all of the corresponding features of the two triangles are going to be congruent. In particular, side DC, side DC corresponds to side BA, side DC on this bottom triangle corresponds to side BA on that top triangle. So they need to be congruent. So DC, so we get DC is going to be equal to BA, and that's because they are corresponding sides, corresponding sides of congruent, congruent triangles. So this is going to be equal to that, and by that exact same logic, AD, AD corresponds to CB. AD corresponds to CB. AD is equal to CB, and for the exact same reason, corresponding sides of congruent triangles.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So DC, so we get DC is going to be equal to BA, and that's because they are corresponding sides, corresponding sides of congruent, congruent triangles. So this is going to be equal to that, and by that exact same logic, AD, AD corresponds to CB. AD corresponds to CB. AD is equal to CB, and for the exact same reason, corresponding sides of congruent triangles. And then we're done. We've proven that opposite sides are congruent. Now let's go the other way.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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AD is equal to CB, and for the exact same reason, corresponding sides of congruent triangles. And then we're done. We've proven that opposite sides are congruent. Now let's go the other way. Let's go the other way. Let's say that we have some type of a quadrilateral, and we know that the opposite sides are congruent. Can we prove to ourselves that this is a parallelogram?
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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Now let's go the other way. Let's go the other way. Let's say that we have some type of a quadrilateral, and we know that the opposite sides are congruent. Can we prove to ourselves that this is a parallelogram? Well, it's kind of the same proof in reverse. So let's draw a diagonal here, since we know a lot about triangles. So let me draw, there we go.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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Can we prove to ourselves that this is a parallelogram? Well, it's kind of the same proof in reverse. So let's draw a diagonal here, since we know a lot about triangles. So let me draw, there we go. That's the hardest part. Let's see, draw it, that's pretty good. All right, so we obviously know that CB is going to be equal to itself.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So let me draw, there we go. That's the hardest part. Let's see, draw it, that's pretty good. All right, so we obviously know that CB is going to be equal to itself. So I'll draw it like that. We all, obviously, because it's the same line. And then we have something interesting.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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All right, so we obviously know that CB is going to be equal to itself. So I'll draw it like that. We all, obviously, because it's the same line. And then we have something interesting. We've split this quadrilateral into two triangles, triangle ACB and triangle DBC. And notice, they have, all three sides of these two triangles are equal to each other. So we know by side, side, side, that they are congruent.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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And then we have something interesting. We've split this quadrilateral into two triangles, triangle ACB and triangle DBC. And notice, they have, all three sides of these two triangles are equal to each other. So we know by side, side, side, that they are congruent. So we know that triangle, triangle A, and we're starting A, and then I'm going to the one-half side. So ACB is congruent to triangle DBC. DBC, and this is by side, side, side, side, side, side, side, congruency.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So we know by side, side, side, that they are congruent. So we know that triangle, triangle A, and we're starting A, and then I'm going to the one-half side. So ACB is congruent to triangle DBC. DBC, and this is by side, side, side, side, side, side, side, congruency. Well, what does that do for us? Well, it tells us that all of the corresponding angles are going to be congruent. So for example, ABC, angle ABC is going to be, so let me mark that.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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DBC, and this is by side, side, side, side, side, side, side, congruency. Well, what does that do for us? Well, it tells us that all of the corresponding angles are going to be congruent. So for example, ABC, angle ABC is going to be, so let me mark that. Angle ABC is going to be congruent, and you can say ABC is going to be congruent to DCB. DCB, angle DCB, and you could say by, you could say corresponding angles congruent of congruent triangles. I'm just using some shorthand here to save some time.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So for example, ABC, angle ABC is going to be, so let me mark that. Angle ABC is going to be congruent, and you can say ABC is going to be congruent to DCB. DCB, angle DCB, and you could say by, you could say corresponding angles congruent of congruent triangles. I'm just using some shorthand here to save some time. So ABC is going to be congruent to DCB. So these two angles are going to be congruent. Well, this is interesting because here you have a line, and it's intersecting AB and CD, and we clearly see that these things that could be alternate angles, alternate interior angles, are congruent.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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I'm just using some shorthand here to save some time. So ABC is going to be congruent to DCB. So these two angles are going to be congruent. Well, this is interesting because here you have a line, and it's intersecting AB and CD, and we clearly see that these things that could be alternate angles, alternate interior angles, are congruent. And because we have these congruent alternate interior angles, we know that AB must be parallel to CD. So this must be parallel to that. So we know that AB is parallel to CD by alternate interior angles of a transversal intersecting parallel lines.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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Well, this is interesting because here you have a line, and it's intersecting AB and CD, and we clearly see that these things that could be alternate angles, alternate interior angles, are congruent. And because we have these congruent alternate interior angles, we know that AB must be parallel to CD. So this must be parallel to that. So we know that AB is parallel to CD by alternate interior angles of a transversal intersecting parallel lines. Now, we can use that exact same logic. We also know that angle, let me get this right, angle ACB is congruent to angle DBC. And we know that by corresponding angles congruent of congruent triangles.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So we know that AB is parallel to CD by alternate interior angles of a transversal intersecting parallel lines. Now, we can use that exact same logic. We also know that angle, let me get this right, angle ACB is congruent to angle DBC. And we know that by corresponding angles congruent of congruent triangles. So we're just saying that this angle is equal to that angle. Well, once again, these could be alternate interior angles. They look like they could be.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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And we know that by corresponding angles congruent of congruent triangles. So we're just saying that this angle is equal to that angle. Well, once again, these could be alternate interior angles. They look like they could be. This is a transversal, and here's two lines here, which we're not sure whether they're parallel. But because the alternate interior angles are congruent, we know that they are parallel. So this is parallel to that.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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They look like they could be. This is a transversal, and here's two lines here, which we're not sure whether they're parallel. But because the alternate interior angles are congruent, we know that they are parallel. So this is parallel to that. So we know that AC is parallel to BD by alternate interior angles. By alternate interior angles. And we're done.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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So this is parallel to that. So we know that AC is parallel to BD by alternate interior angles. By alternate interior angles. And we're done. So what we've done is, it's interesting, we've shown if you have a parallelogram, opposite sides have the same length. And if opposite sides have the same length, then you have a parallelogram. And so we've actually proven it in both directions.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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And we're done. So what we've done is, it's interesting, we've shown if you have a parallelogram, opposite sides have the same length. And if opposite sides have the same length, then you have a parallelogram. And so we've actually proven it in both directions. And so we can actually make what you call an if and only if statement. You can say if opposite sides are parallel of a quadrilateral, or you could say opposite sides of a quadrilateral are parallel if and only if their lengths are equal. And you say if and only if.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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And so we've actually proven it in both directions. And so we can actually make what you call an if and only if statement. You can say if opposite sides are parallel of a quadrilateral, or you could say opposite sides of a quadrilateral are parallel if and only if their lengths are equal. And you say if and only if. So if they are parallel, then you could say their lengths are equal. And only if, only if their lengths are equal are they parallel. We've proven it in both directions.
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Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3
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You didn't have to be told it's a hexagon. But the regular part lets us know that all of the sides, all six sides have the same length and all of the interior angles have the same measure. Fair enough. And then they give us the length of one of the sides. And since this is a regular hexagon, they're actually giving us the length of all of the sides. They say it's two squares of three. So this side right over here is two squares of three.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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And then they give us the length of one of the sides. And since this is a regular hexagon, they're actually giving us the length of all of the sides. They say it's two squares of three. So this side right over here is two squares of three. This side over here is two squares of three. And I could just go around the hexagon. Every one of their sides is two squares of three.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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So this side right over here is two squares of three. This side over here is two squares of three. And I could just go around the hexagon. Every one of their sides is two squares of three. They want us to find the area of this hexagon. Find the area of A, B, C, D, E, F. And the best way to find the area, especially of regular polygons, is try to split it up into triangles. And hexagons are a bit of a special case.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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Every one of their sides is two squares of three. They want us to find the area of this hexagon. Find the area of A, B, C, D, E, F. And the best way to find the area, especially of regular polygons, is try to split it up into triangles. And hexagons are a bit of a special case. Maybe in future videos we'll think about the more general case of any polygon. But with a hexagon, what you could think about is if we take this point right over here, and let's call this point G, and let's say it's the center of the hexagon. And when I'm talking about a center of a hexagon, I'm talking about a point.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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And hexagons are a bit of a special case. Maybe in future videos we'll think about the more general case of any polygon. But with a hexagon, what you could think about is if we take this point right over here, and let's call this point G, and let's say it's the center of the hexagon. And when I'm talking about a center of a hexagon, I'm talking about a point. It can't be equidistant from everything over here because this isn't a circle. But we could say it's equidistant from all of the vertices. So GD is the same thing as GC, is the same thing as GB, which is the same thing as GA, which is the same thing as GF, which is the same thing as GE.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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And when I'm talking about a center of a hexagon, I'm talking about a point. It can't be equidistant from everything over here because this isn't a circle. But we could say it's equidistant from all of the vertices. So GD is the same thing as GC, is the same thing as GB, which is the same thing as GA, which is the same thing as GF, which is the same thing as GE. So let me draw some of those that I just talked about. So that is GE. There's GD.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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So GD is the same thing as GC, is the same thing as GB, which is the same thing as GA, which is the same thing as GF, which is the same thing as GE. So let me draw some of those that I just talked about. So that is GE. There's GD. There's GC. All of these lengths are going to be the same. So there's a point G, which we can call the center, the center of this polygon.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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There's GD. There's GC. All of these lengths are going to be the same. So there's a point G, which we can call the center, the center of this polygon. And we know that this length is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length. We also know that if we go all the way around the circle, if we go all the way around a circle like that, we've gone 360 degrees. And we know that these triangles, these triangles are all going to be congruent to each other.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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So there's a point G, which we can call the center, the center of this polygon. And we know that this length is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length. We also know that if we go all the way around the circle, if we go all the way around a circle like that, we've gone 360 degrees. And we know that these triangles, these triangles are all going to be congruent to each other. And there's multiple ways that we could show it. But the easiest way is, look, they have two sides. All of them have this side and this side be congruent to each other because G is in the center.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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And we know that these triangles, these triangles are all going to be congruent to each other. And there's multiple ways that we could show it. But the easiest way is, look, they have two sides. All of them have this side and this side be congruent to each other because G is in the center. And they all have this third common side of 2 squared is 3. So all of them by side, side, side, they are all congruent. What that tells us is if they're all congruent, then this angle, this interior angle right over here, is going to be the same for all 6 of these triangles over here.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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All of them have this side and this side be congruent to each other because G is in the center. And they all have this third common side of 2 squared is 3. So all of them by side, side, side, they are all congruent. What that tells us is if they're all congruent, then this angle, this interior angle right over here, is going to be the same for all 6 of these triangles over here. And maybe I call that x. That's angle x. That's x.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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What that tells us is if they're all congruent, then this angle, this interior angle right over here, is going to be the same for all 6 of these triangles over here. And maybe I call that x. That's angle x. That's x. That's x. That's x. That's x.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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That's x. That's x. That's x. That's x. And if you add them all up, we've gone around a circle. We've gone 360 degrees. And we have 6 of these x's.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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That's x. And if you add them all up, we've gone around a circle. We've gone 360 degrees. And we have 6 of these x's. So you get 6x is equal to 360 degrees. You divide both sides by 6. You get x is equal to 60 degrees.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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And we have 6 of these x's. So you get 6x is equal to 360 degrees. You divide both sides by 6. You get x is equal to 60 degrees. All of these are equal to 60 degrees. Now there's something interesting. We know that these triangles, for example, triangle GBC, and we could do that for any of these 6 triangles.
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Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3
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