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So they're telling us that line AOB. And they could have just said line AB, but I guess they wanted to put the O in there to show that point O is on that line, that AOB are collinear. And COD is our straight lines. All right, fair enough. Which of these statements prove vertical angles are always equal? So vertical angles would be the angles on the opposite sides of an intersection. So in order to prove that vertical, so for example, angle AOC and angle DOB are vertical angles.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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All right, fair enough. Which of these statements prove vertical angles are always equal? So vertical angles would be the angles on the opposite sides of an intersection. So in order to prove that vertical, so for example, angle AOC and angle DOB are vertical angles. And if we wanted to prove that they are equal, we would say, well, their measures are going to be equal. So theta should be equal to phi. So let's see which of these statements actually does that.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So in order to prove that vertical, so for example, angle AOC and angle DOB are vertical angles. And if we wanted to prove that they are equal, we would say, well, their measures are going to be equal. So theta should be equal to phi. So let's see which of these statements actually does that. So this one says segment OA is congruent to OD. We don't know that. They never even told us that.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So let's see which of these statements actually does that. So this one says segment OA is congruent to OD. We don't know that. They never even told us that. So I don't even have to read the rest of it. This is already saying, I don't know how far D is away from O. I don't know if it's the same distance as A is from O. So we can just rule this first choice out.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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They never even told us that. So I don't even have to read the rest of it. This is already saying, I don't know how far D is away from O. I don't know if it's the same distance as A is from O. So we can just rule this first choice out. I can just stop reading. They started with a statement that we don't know based on the information they gave us. So let's look at the second choice.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So we can just rule this first choice out. I can just stop reading. They started with a statement that we don't know based on the information they gave us. So let's look at the second choice. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to OB and OD respectively. If two rays are rotated by the same amount, the angle between them will not change. So phi must be equal to theta.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So let's look at the second choice. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to OB and OD respectively. If two rays are rotated by the same amount, the angle between them will not change. So phi must be equal to theta. So this is interesting. So let's just slow down and think about what they're saying. If ray OA and OC are each rotated 180 degrees, so if you take ray OA, this ray right over here, if you rotate it 180 degrees, it's going to go all the way around and point in the other direction.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So phi must be equal to theta. So this is interesting. So let's just slow down and think about what they're saying. If ray OA and OC are each rotated 180 degrees, so if you take ray OA, this ray right over here, if you rotate it 180 degrees, it's going to go all the way around and point in the other direction. So it's going to map to ray OB. So I definitely believe that. OA is going to map to ray OB.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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If ray OA and OC are each rotated 180 degrees, so if you take ray OA, this ray right over here, if you rotate it 180 degrees, it's going to go all the way around and point in the other direction. So it's going to map to ray OB. So I definitely believe that. OA is going to map to ray OB. And OC, ray OC, if you rotate 180 degrees, is going to map to ray OD. And so this first statement is true. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to ray OB and OD respectively.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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OA is going to map to ray OB. And OC, ray OC, if you rotate 180 degrees, is going to map to ray OD. And so this first statement is true. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to ray OB and OD respectively. And when people say respectively, they're saying in the same order, that ray OA maps to ray OB and that ray OC maps to ray OD. And we saw that. Ray OA maps, if you rotate it all the way around 180 degrees, it'll map to OB.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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If ray OA and ray OC are each rotated 180 degrees about point O, they must map to ray OB and OD respectively. And when people say respectively, they're saying in the same order, that ray OA maps to ray OB and that ray OC maps to ray OD. And we saw that. Ray OA maps, if you rotate it all the way around 180 degrees, it'll map to OB. And OC, if you rotate 180 degrees, will map to OD. So I'm feeling good about that first sentence. If two rays are rotated by the same amount, the angle between them will not change.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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Ray OA maps, if you rotate it all the way around 180 degrees, it'll map to OB. And OC, if you rotate 180 degrees, will map to OD. So I'm feeling good about that first sentence. If two rays are rotated by the same amount, the angle between them will not change. Especially if they are rotated around, yeah, I'll go with that. If two rays are rotated by the same amount, the angle between them will not change. So if we rotate both of these rays by 180 degrees, then we've essentially mapped to OB and OD.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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If two rays are rotated by the same amount, the angle between them will not change. Especially if they are rotated around, yeah, I'll go with that. If two rays are rotated by the same amount, the angle between them will not change. So if we rotate both of these rays by 180 degrees, then we've essentially mapped to OB and OD. Or another way to think about it, this angle, angle AOC, is going to map to angle BOD. And so the measure of those angles are going to be the same. So phi must be equal to theta.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So if we rotate both of these rays by 180 degrees, then we've essentially mapped to OB and OD. Or another way to think about it, this angle, angle AOC, is going to map to angle BOD. And so the measure of those angles are going to be the same. So phi must be equal to theta. So I actually like this second statement a lot. So let's see this last statement. Rotations preserve lengths and angles.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So phi must be equal to theta. So I actually like this second statement a lot. So let's see this last statement. Rotations preserve lengths and angles. AB is congruent to CD. Actually, we don't know whether segment AB is congruent to CD. They never told us that.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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Rotations preserve lengths and angles. AB is congruent to CD. Actually, we don't know whether segment AB is congruent to CD. They never told us that. We don't know how far apart these things are. So we know that phi is equal to theta. So this statement right over here is just suspect.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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They never told us that. We don't know how far apart these things are. So we know that phi is equal to theta. So this statement right over here is just suspect. And so actually, I don't like that one. So I'm going to go with the first one, which is it takes a little bit of visualization going on. But if you took angle AOC and you rotated it 180 degrees, which means take the corresponding rays, or the rays that make it up, and rotate them 180 degrees, you get to angle BOD.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So we have this larger triangle here, and inside of that we have these other triangles. And we're given this information right over here, that triangle BCD is congruent to triangle BCA, which is congruent to triangle ECD. And given just this information, what I want to do in this drawing, I want to figure out what every angle on this drawing is. What's the measure of every angle? So let's see what we can do here. So let's just start with the information that they've actually given us. So we know that triangle BCD is congruent to, well, we know all of these three triangles are congruent to each other.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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What's the measure of every angle? So let's see what we can do here. So let's just start with the information that they've actually given us. So we know that triangle BCD is congruent to, well, we know all of these three triangles are congruent to each other. So for example, BCD is congruent to ECD. And so their corresponding sides and corresponding angles will also be congruent. So just looking at the order in which they're written, vertex B corresponds in this triangle, in BCD, corresponds to vertex B in BCA, so this is the B vertex in BCA, which corresponds to the E vertex in ECD.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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So we know that triangle BCD is congruent to, well, we know all of these three triangles are congruent to each other. So for example, BCD is congruent to ECD. And so their corresponding sides and corresponding angles will also be congruent. So just looking at the order in which they're written, vertex B corresponds in this triangle, in BCD, corresponds to vertex B in BCA, so this is the B vertex in BCA, which corresponds to the E vertex in ECD. So everything that I've done in magenta, all of these angles are congruent. And then we also know that the C angle, so in BCA, sorry, BCD, this angle, this angle right over here, is congruent to the C angle in BCA. BCA, the C angle is right over here, or C is the vertex for that angle in BCA.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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So just looking at the order in which they're written, vertex B corresponds in this triangle, in BCD, corresponds to vertex B in BCA, so this is the B vertex in BCA, which corresponds to the E vertex in ECD. So everything that I've done in magenta, all of these angles are congruent. And then we also know that the C angle, so in BCA, sorry, BCD, this angle, this angle right over here, is congruent to the C angle in BCA. BCA, the C angle is right over here, or C is the vertex for that angle in BCA. And that is also the C angle, I guess we could call it, in ECD. But in ECD, we're talking about this angle right over here. So these three angles are going to be congruent.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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BCA, the C angle is right over here, or C is the vertex for that angle in BCA. And that is also the C angle, I guess we could call it, in ECD. But in ECD, we're talking about this angle right over here. So these three angles are going to be congruent. And I think you could already guess a way to come up with the values of those three angles. But let's just keep looking at everything else that they're telling us. Finally, we have vertex D over here.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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So these three angles are going to be congruent. And I think you could already guess a way to come up with the values of those three angles. But let's just keep looking at everything else that they're telling us. Finally, we have vertex D over here. So angle, so this is the last one where we listed B. So in triangle BCD, this angle right over here corresponds to the A vertex angle in BCA. So BCA, that's going to correspond to this angle right over here.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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Finally, we have vertex D over here. So angle, so this is the last one where we listed B. So in triangle BCD, this angle right over here corresponds to the A vertex angle in BCA. So BCA, that's going to correspond to this angle right over here. It's really the only one that we haven't labeled yet. And that corresponds to this angle, this vertex right over here, that angle right over there. And just to make it consistent, this C should also be circled in yellow.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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So BCA, that's going to correspond to this angle right over here. It's really the only one that we haven't labeled yet. And that corresponds to this angle, this vertex right over here, that angle right over there. And just to make it consistent, this C should also be circled in yellow. And so we have all these congruencies, and now we can come up with some interesting things about them. First of all, here, angle BCA, angle BCD, and angle DCE, they're all congruent. And when you add them up together, you get to 180 degrees.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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And just to make it consistent, this C should also be circled in yellow. And so we have all these congruencies, and now we can come up with some interesting things about them. First of all, here, angle BCA, angle BCD, and angle DCE, they're all congruent. And when you add them up together, you get to 180 degrees. If you put them all adjacent, as they all are right here, they end up with a straight angle if you look at their outer sides. So if these are each x, you have three of them added together have to be 180 degrees, which tells us that each of these have to be 60 degrees. That's the only way you have three of the same thing adding up to 180 degrees.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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And when you add them up together, you get to 180 degrees. If you put them all adjacent, as they all are right here, they end up with a straight angle if you look at their outer sides. So if these are each x, you have three of them added together have to be 180 degrees, which tells us that each of these have to be 60 degrees. That's the only way you have three of the same thing adding up to 180 degrees. Fair enough. What else can we do? Well, we have these two characters up here.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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That's the only way you have three of the same thing adding up to 180 degrees. Fair enough. What else can we do? Well, we have these two characters up here. They are both equal, and they add up to 180 degrees. They are supplementary. And the only way you can have two equal things that add up to 180 is if they're both 90 degrees.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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Well, we have these two characters up here. They are both equal, and they add up to 180 degrees. They are supplementary. And the only way you can have two equal things that add up to 180 is if they're both 90 degrees. So these two characters are both 90 degrees, or we could say this is a right angle, that's a right angle. And this is congruent to both of those, so that is also 90 degrees. And then we're left with these magenta parts of the angle.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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And the only way you can have two equal things that add up to 180 is if they're both 90 degrees. So these two characters are both 90 degrees, or we could say this is a right angle, that's a right angle. And this is congruent to both of those, so that is also 90 degrees. And then we're left with these magenta parts of the angle. And here we can just say, well, 90 plus 60 plus something is going to add up to 180. 90 plus 60 is 150, so this has to be 30 degrees to add up to 180. And if that's 30 degrees, then this is 30 degrees.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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And then we're left with these magenta parts of the angle. And here we can just say, well, 90 plus 60 plus something is going to add up to 180. 90 plus 60 is 150, so this has to be 30 degrees to add up to 180. And if that's 30 degrees, then this is 30 degrees. And then this thing right over here is 30 degrees. And then the last thing, we've actually done what we said we would do. We found out all of the angles.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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And if that's 30 degrees, then this is 30 degrees. And then this thing right over here is 30 degrees. And then the last thing, we've actually done what we said we would do. We found out all of the angles. We can also think about these outer angles. Or not the outer angles, but these combined angles. So angle, say, angle ABE.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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We found out all of the angles. We can also think about these outer angles. Or not the outer angles, but these combined angles. So angle, say, angle ABE. So this whole angle we see is 60 degrees. This angle is 90 degrees. And this angle here is 30.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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So angle, say, angle ABE. So this whole angle we see is 60 degrees. This angle is 90 degrees. And this angle here is 30. So what's interesting is these three smaller triangles, they all have the exact same angles, 30, 60, 90, and the exact same side lengths. And we know that because they're congruent. And what's interesting is when you put them together this way, they construct this larger triangle, triangle ABE, that's clearly not congruent.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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And this angle here is 30. So what's interesting is these three smaller triangles, they all have the exact same angles, 30, 60, 90, and the exact same side lengths. And we know that because they're congruent. And what's interesting is when you put them together this way, they construct this larger triangle, triangle ABE, that's clearly not congruent. It's a larger triangle. It has different measures for its length. But it has the same angles, 30, 60, and then 90.
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Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3
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I know what you're thinking, what's all of the silliness on the right-hand side? This is actually just the view we use when we're trying to debug things on Khan Academy, but we can still do the exercise. So it says drag the center point and perimeter of the circle to graph the equation. So the first thing we want to think about is, well, what's the center of this equation? Well, a standard form of a circle is x minus the x-coordinate of the center squared plus y minus the y-coordinate of the center squared is equal to the radius squared. So x minus the x-coordinate of the center, so the x-coordinate of the center must be negative five. Cause the way we can get a positive 5 here is by subtracting a negative 5.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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So the first thing we want to think about is, well, what's the center of this equation? Well, a standard form of a circle is x minus the x-coordinate of the center squared plus y minus the y-coordinate of the center squared is equal to the radius squared. So x minus the x-coordinate of the center, so the x-coordinate of the center must be negative five. Cause the way we can get a positive 5 here is by subtracting a negative 5. So the x-coordinate must be negative 5 and the y-coordinate must be positive 5 cause y minus the y-coordinate of the center. So y-coordinate is positive five. And then the radius squared is going to be equal to four.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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Cause the way we can get a positive 5 here is by subtracting a negative 5. So the x-coordinate must be negative 5 and the y-coordinate must be positive 5 cause y minus the y-coordinate of the center. So y-coordinate is positive five. And then the radius squared is going to be equal to four. So that means that the radius is equal to two. And the way it's drawn right now, I mean we could drag this out like this, but this, the way it's drawn, the radius is indeed equal to two. And so we are done.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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And then the radius squared is going to be equal to four. So that means that the radius is equal to two. And the way it's drawn right now, I mean we could drag this out like this, but this, the way it's drawn, the radius is indeed equal to two. And so we are done. And I really want to hit the point home of what I just did. So let me get my little, get my scratch pad out. So this, sorry for knocking the microphone just now.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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And so we are done. And I really want to hit the point home of what I just did. So let me get my little, get my scratch pad out. So this, sorry for knocking the microphone just now. So that equation was x plus five squared plus y minus five squared is equal to four squared. And so I want to rewrite this as, this is x minus negative five, x minus negative five squared plus y minus positive five, positive five squared is equal to, and instead of writing it as four, I'll write it as two squared. And so this right over here tells us that the center of the circle is going to be, x equals negative five, y equals five, and the radius is going to be equal to two.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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So this, sorry for knocking the microphone just now. So that equation was x plus five squared plus y minus five squared is equal to four squared. And so I want to rewrite this as, this is x minus negative five, x minus negative five squared plus y minus positive five, positive five squared is equal to, and instead of writing it as four, I'll write it as two squared. And so this right over here tells us that the center of the circle is going to be, x equals negative five, y equals five, and the radius is going to be equal to two. And once again, this is no magic here. This is not just, I don't want you to just memorize this formula. I want you to appreciate that this formula comes straight out of the Pythagorean theorem, straight out of the distance formula, which comes out of the Pythagorean theorem.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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And so this right over here tells us that the center of the circle is going to be, x equals negative five, y equals five, and the radius is going to be equal to two. And once again, this is no magic here. This is not just, I don't want you to just memorize this formula. I want you to appreciate that this formula comes straight out of the Pythagorean theorem, straight out of the distance formula, which comes out of the Pythagorean theorem. Remember, if you have some center, in this case it's the point negative five comma five. So negative five comma five. And you want to find all of the x's and y's that are two away from it.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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I want you to appreciate that this formula comes straight out of the Pythagorean theorem, straight out of the distance formula, which comes out of the Pythagorean theorem. Remember, if you have some center, in this case it's the point negative five comma five. So negative five comma five. And you want to find all of the x's and y's that are two away from it. So you want to find all the x's and y's that are two away from it. So that would be one of them. X comma y.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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And you want to find all of the x's and y's that are two away from it. So you want to find all the x's and y's that are two away from it. So that would be one of them. X comma y. This distance is two. And there's going to be a bunch of them. And when you plot all of them together, you're going to get a circle with a radius two around that center.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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X comma y. This distance is two. And there's going to be a bunch of them. And when you plot all of them together, you're going to get a circle with a radius two around that center. But let's think about how we got that actual formula. Well, the distance between that coordinate, between any of these x's and y's, it could be an x and y here, it could be an x and y here, and this is going to be two. So we could have our change in x.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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And when you plot all of them together, you're going to get a circle with a radius two around that center. But let's think about how we got that actual formula. Well, the distance between that coordinate, between any of these x's and y's, it could be an x and y here, it could be an x and y here, and this is going to be two. So we could have our change in x. So we have x minus negative five. So that's our change in x between any point x comma y and negative five comma five. So our change in x squared plus our change in y squared, so that's going to be y minus the y coordinate over here, squared, is going to be equal to the radius squared.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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So we could have our change in x. So we have x minus negative five. So that's our change in x between any point x comma y and negative five comma five. So our change in x squared plus our change in y squared, so that's going to be y minus the y coordinate over here, squared, is going to be equal to the radius squared. So the change in y, it's going to be from this y to that y. If this is the end point, it'd be the end minus the beginning, y minus five. Y minus five squared.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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So our change in x squared plus our change in y squared, so that's going to be y minus the y coordinate over here, squared, is going to be equal to the radius squared. So the change in y, it's going to be from this y to that y. If this is the end point, it'd be the end minus the beginning, y minus five. Y minus five squared. And so this shows for any xy that is two away from the center, this equation will hold. And it becomes, I'll just write this in neutral color, x plus five squared plus y minus five squared is equal to the radius squared, is equal to two, or let me just write that, is equal to four. And let me make it, I really want to, you know, I dislike it when formulas are just memorized and you don't see the connection to other things.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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Y minus five squared. And so this shows for any xy that is two away from the center, this equation will hold. And it becomes, I'll just write this in neutral color, x plus five squared plus y minus five squared is equal to the radius squared, is equal to two, or let me just write that, is equal to four. And let me make it, I really want to, you know, I dislike it when formulas are just memorized and you don't see the connection to other things. Notice, we can construct, we can construct a nice little right triangle here. So our change in x is that right over there. So that is our change in x, change in x.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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And let me make it, I really want to, you know, I dislike it when formulas are just memorized and you don't see the connection to other things. Notice, we can construct, we can construct a nice little right triangle here. So our change in x is that right over there. So that is our change in x, change in x. And our change in y, our change in y, not the change in y squared, but our change in y is that right over there. Change in y, our change in y, you could view that as y minus, so this is change in y is going to be y minus five. And our change in x is x minus negative five.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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So that is our change in x, change in x. And our change in y, our change in y, not the change in y squared, but our change in y is that right over there. Change in y, our change in y, you could view that as y minus, so this is change in y is going to be y minus five. And our change in x is x minus negative five. X minus negative five. So this is just change in x squared plus change in y squared is equal to the hypotenuse squared, which is the length of, which is this radius. So once again, comes straight out of the Pythagorean theorem.
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Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3
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And I think you know why they're called this. The measures of its angles are 30 degrees, 60 degrees, and 90 degrees. And what we're going to prove in this video, and this tends to be a very useful result, at least for a lot of what you see in a geometry class and then later on in a trigonometry class, is the ratios between the sides of a 30-60-90 triangle. That if the hypotenuse has length x, so remember the hypotenuse is opposite the 90-degree side, if the hypotenuse has length x, what we're going to prove is that the shortest side, which is opposite the 30-degree side, has length x over 2, and that the 60-degree side, or the side that's opposite the 60-degree angle, I should say, is going to be square root of 3 times the shortest side. So square root of 3 times x over 2. That's going to be its length. So that's what we're going to prove in this video.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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That if the hypotenuse has length x, so remember the hypotenuse is opposite the 90-degree side, if the hypotenuse has length x, what we're going to prove is that the shortest side, which is opposite the 30-degree side, has length x over 2, and that the 60-degree side, or the side that's opposite the 60-degree angle, I should say, is going to be square root of 3 times the shortest side. So square root of 3 times x over 2. That's going to be its length. So that's what we're going to prove in this video. And then in other videos we're going to apply this. We're going to show that this is actually a pretty useful result. Now let's start with a triangle that we're very familiar with.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So that's what we're going to prove in this video. And then in other videos we're going to apply this. We're going to show that this is actually a pretty useful result. Now let's start with a triangle that we're very familiar with. So let me draw ourselves an equilateral triangle. So drawing the triangles is always the hard part. So this is my best shot at an equilateral triangle.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Now let's start with a triangle that we're very familiar with. So let me draw ourselves an equilateral triangle. So drawing the triangles is always the hard part. So this is my best shot at an equilateral triangle. And let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So this is my best shot at an equilateral triangle. And let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say that it has length, and let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say that it has length, and let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now what I'm going to do is I'm going to drop an altitude from this top point right over here.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now what I'm going to do is I'm going to drop an altitude from this top point right over here. So I'm going to drop an altitude right down, and by definition when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that this is not only an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Now what I'm going to do is I'm going to drop an altitude from this top point right over here. So I'm going to drop an altitude right down, and by definition when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that this is not only an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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And it's a pretty straightforward proof to show that this is not only an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. And then we could do, we have this angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other, then the third angle has to be congruent to each other.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So this side is common to both of them right over here. And then we could do, we have this angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other, then the third angle has to be congruent to each other. So this angle right over here needs to be congruent to that angle right over there. So these two are congruent. You can use actually a variety of our congruence postulates.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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And so if these two are congruent to each other, then the third angle has to be congruent to each other. So this angle right over here needs to be congruent to that angle right over there. So these two are congruent. You can use actually a variety of our congruence postulates. We could say side, angle, side, side, angle, side congruence. We could use angle, side, angle, any of those to show that triangle A, B, D is congruent to triangle C, B, D. And what that does for us, and we could use, as I said, we could use angle, side, angle, or side, angle, side, whatever we like to use for this. What that does for us is it tells us that the corresponding sides of these triangles are going to be equal.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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You can use actually a variety of our congruence postulates. We could say side, angle, side, side, angle, side congruence. We could use angle, side, angle, any of those to show that triangle A, B, D is congruent to triangle C, B, D. And what that does for us, and we could use, as I said, we could use angle, side, angle, or side, angle, side, whatever we like to use for this. What that does for us is it tells us that the corresponding sides of these triangles are going to be equal. In particular, the length of, in particular, A, D is going to be equal to C, D. These are corresponding sides. So these are going to be equal to each other. And if we know that they're equal to each other and they add up to x, remember this was an equilateral triangle of length x, we know that this side right over here is going to be x over 2.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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What that does for us is it tells us that the corresponding sides of these triangles are going to be equal. In particular, the length of, in particular, A, D is going to be equal to C, D. These are corresponding sides. So these are going to be equal to each other. And if we know that they're equal to each other and they add up to x, remember this was an equilateral triangle of length x, we know that this side right over here is going to be x over 2. We know this is going to be x over 2. Not only do we know that, but we also knew when we dropped this altitude, we showed that this angle has to be congruent to that angle, and their measures have to add up to 60. So if two things are the same and they add up to 60, this is going to be at 30 degrees, and this is going to be 30 degrees.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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And if we know that they're equal to each other and they add up to x, remember this was an equilateral triangle of length x, we know that this side right over here is going to be x over 2. We know this is going to be x over 2. Not only do we know that, but we also knew when we dropped this altitude, we showed that this angle has to be congruent to that angle, and their measures have to add up to 60. So if two things are the same and they add up to 60, this is going to be at 30 degrees, and this is going to be 30 degrees. So we've already shown one of the interesting parts of a 30-60-90 triangle that if the hypotenuse, notice, and I guess I didn't point this out, by dropping this altitude, but I've essentially split this equilateral triangle into two 30-60-90 triangles. And so we've already shown that if the side opposite the 90 degree side is x, that the side opposite the 30 degree side is going to be x over 2. That's what we showed right over here.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So if two things are the same and they add up to 60, this is going to be at 30 degrees, and this is going to be 30 degrees. So we've already shown one of the interesting parts of a 30-60-90 triangle that if the hypotenuse, notice, and I guess I didn't point this out, by dropping this altitude, but I've essentially split this equilateral triangle into two 30-60-90 triangles. And so we've already shown that if the side opposite the 90 degree side is x, that the side opposite the 30 degree side is going to be x over 2. That's what we showed right over here. Now we just have to come up with the third side, the side that is opposite the 60 degree side right over there. And let's call that length, well, I'll just use the letters that we already have here. This is BD.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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That's what we showed right over here. Now we just have to come up with the third side, the side that is opposite the 60 degree side right over there. And let's call that length, well, I'll just use the letters that we already have here. This is BD. And we can just use the Pythagorean theorem right here. BD squared plus this length right over here squared plus x over 2 squared is going to be equal to the hypotenuse squared. So we get BD squared plus x over 2 squared.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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This is BD. And we can just use the Pythagorean theorem right here. BD squared plus this length right over here squared plus x over 2 squared is going to be equal to the hypotenuse squared. So we get BD squared plus x over 2 squared. This is just straight out of the Pythagorean theorem. Plus x over 2 squared is going to equal this hypotenuse squared is going to equal x squared. And just to be clear, I'm looking at this triangle right here.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So we get BD squared plus x over 2 squared. This is just straight out of the Pythagorean theorem. Plus x over 2 squared is going to equal this hypotenuse squared is going to equal x squared. And just to be clear, I'm looking at this triangle right here. I'm looking at this triangle right over here on the right. And I'm just applying the Pythagorean theorem. This side squared plus this side squared is going to equal the hypotenuse squared.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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And just to be clear, I'm looking at this triangle right here. I'm looking at this triangle right over here on the right. And I'm just applying the Pythagorean theorem. This side squared plus this side squared is going to equal the hypotenuse squared. And let's solve now for BD. You get BD squared plus x squared over 4 is equal to x squared. You could view this as 4x squared over 4.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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This side squared plus this side squared is going to equal the hypotenuse squared. And let's solve now for BD. You get BD squared plus x squared over 4 is equal to x squared. You could view this as 4x squared over 4. That's the same thing, obviously, as x squared. And then if you subtract 1 fourth x squared from both sides or x squared over 4 from both sides, you get BD squared is equal to 4x squared over 4 minus x squared over 4 is going to be 3x squared over 4. So we're just going to be 3x squared over 4.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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You could view this as 4x squared over 4. That's the same thing, obviously, as x squared. And then if you subtract 1 fourth x squared from both sides or x squared over 4 from both sides, you get BD squared is equal to 4x squared over 4 minus x squared over 4 is going to be 3x squared over 4. So we're just going to be 3x squared over 4. Take the principal root of both sides. You get BD is equal to the square root of 3 times x. Principal root of 3 is square root of 3.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So we're just going to be 3x squared over 4. Take the principal root of both sides. You get BD is equal to the square root of 3 times x. Principal root of 3 is square root of 3. Principal root of x squared is just x over the principal root of 4, which is 2. And BD is the side opposite the 60 degree side. So we're done.
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30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Point A is at negative 5 comma 5. So this is negative 5 right over here. This is 1, 2, 3, 4, 5. That's 5 right over there. So point A is right about there. So that is point A, just like that, at negative 5 comma 5. And then it's the center of circle A, which I won't draw just yet because I don't know the radius of circle A.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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That's 5 right over there. So point A is right about there. So that is point A, just like that, at negative 5 comma 5. And then it's the center of circle A, which I won't draw just yet because I don't know the radius of circle A. Point B is at, let me underline these in the appropriate color, point B is at 3 comma 1. So 1, 2, 3 comma 1. So that's point B right over there.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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And then it's the center of circle A, which I won't draw just yet because I don't know the radius of circle A. Point B is at, let me underline these in the appropriate color, point B is at 3 comma 1. So 1, 2, 3 comma 1. So that's point B right over there. It's the center of circle B. Point P is at 0, 0. So it's right over there at the origin.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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So that's point B right over there. It's the center of circle B. Point P is at 0, 0. So it's right over there at the origin. And it is on circles A and B. Well, that's a big piece of information because that tells us if this is on both circles, then that means that this is B's radius away from the point B, from the center. And this tells us that it is circle A's radius away from its center, which is at point A.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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So it's right over there at the origin. And it is on circles A and B. Well, that's a big piece of information because that tells us if this is on both circles, then that means that this is B's radius away from the point B, from the center. And this tells us that it is circle A's radius away from its center, which is at point A. So let's figure out what those radii actually are. And so we can imagine, let me draw a radius, or the radius for circle A. We now know since P sits on it that this could be considered the radius for circle A.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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And this tells us that it is circle A's radius away from its center, which is at point A. So let's figure out what those radii actually are. And so we can imagine, let me draw a radius, or the radius for circle A. We now know since P sits on it that this could be considered the radius for circle A. And you could use a distance formula. But what we'll see is that the distance formula is really just falling out of the Pythagorean theorem. So the distance formula tells us the radius right over here, this is just the distance between those two points.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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We now know since P sits on it that this could be considered the radius for circle A. And you could use a distance formula. But what we'll see is that the distance formula is really just falling out of the Pythagorean theorem. So the distance formula tells us the radius right over here, this is just the distance between those two points. So the radius, or the distance between those two points squared, is going to be equal to our change in x values between A and P. So our change in x values, we could write it as negative 5 minus 0 squared. Negative 5 minus 0 squared, that's our change in x. Negative 5 minus 0 squared plus our change in y.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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So the distance formula tells us the radius right over here, this is just the distance between those two points. So the radius, or the distance between those two points squared, is going to be equal to our change in x values between A and P. So our change in x values, we could write it as negative 5 minus 0 squared. Negative 5 minus 0 squared, that's our change in x. Negative 5 minus 0 squared plus our change in y. 5 minus 0 squared, which gets us that our distance between these two points, which is the length of the radius, squared, is equal to negative 5 squared plus 5 squared. Or we could say that the radius is equal to the square root of, this is 25, this is 25, 50. 50 we can write as 25 times 2.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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Negative 5 minus 0 squared plus our change in y. 5 minus 0 squared, which gets us that our distance between these two points, which is the length of the radius, squared, is equal to negative 5 squared plus 5 squared. Or we could say that the radius is equal to the square root of, this is 25, this is 25, 50. 50 we can write as 25 times 2. So this is equal to the square root of 25 times the square root of 2, which is 5 times the square root of 2. So this distance right over here is 5 times the square root of 2. Now I said this is just the same thing as the Pythagorean theorem.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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50 we can write as 25 times 2. So this is equal to the square root of 25 times the square root of 2, which is 5 times the square root of 2. So this distance right over here is 5 times the square root of 2. Now I said this is just the same thing as the Pythagorean theorem. Why? Well, if we were to construct a right triangle right over here, then we can look at this distance. This distance would be the absolute value of negative 5 minus 0.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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Now I said this is just the same thing as the Pythagorean theorem. Why? Well, if we were to construct a right triangle right over here, then we can look at this distance. This distance would be the absolute value of negative 5 minus 0. Or you could say it's 0 minus negative 5. This distance right over here is 5. This distance is the distance between 0 and 5 in the y direction.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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This distance would be the absolute value of negative 5 minus 0. Or you could say it's 0 minus negative 5. This distance right over here is 5. This distance is the distance between 0 and 5 in the y direction. That's 5. Pythagorean theorem tells us that 5 squared, which is 25, plus 5 squared, another 25, is going to be equal to your hypotenuse squared. And that's exactly what we have here.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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This distance is the distance between 0 and 5 in the y direction. That's 5. Pythagorean theorem tells us that 5 squared, which is 25, plus 5 squared, another 25, is going to be equal to your hypotenuse squared. And that's exactly what we have here. Now you might be saying, wait, wait, wait. This thing had a negative 5 squared here, while here you had a positive 5. But the reason why we could do this is when you square it, the negative disappears.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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And that's exactly what we have here. Now you might be saying, wait, wait, wait. This thing had a negative 5 squared here, while here you had a positive 5. But the reason why we could do this is when you square it, the negative disappears. The distance formula, you could write it this way, where you're taking the absolute value. And then it becomes very clear that this really is just the Pythagorean theorem. This would be 5 squared plus 5 squared.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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But the reason why we could do this is when you square it, the negative disappears. The distance formula, you could write it this way, where you're taking the absolute value. And then it becomes very clear that this really is just the Pythagorean theorem. This would be 5 squared plus 5 squared. 5 squared plus 5 squared. The reason why you don't have to do this is because a sign doesn't matter when you square it. It will become a positive value.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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This would be 5 squared plus 5 squared. 5 squared plus 5 squared. The reason why you don't have to do this is because a sign doesn't matter when you square it. It will become a positive value. But either way, we figured out this radius. Now let's figure out the radius of circle B. The radius of circle B.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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It will become a positive value. But either way, we figured out this radius. Now let's figure out the radius of circle B. The radius of circle B. The same exact idea. The radius of circle B squared is equal to our change in x. So we could write it as 3 minus 0 or 0 minus 3.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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The radius of circle B. The same exact idea. The radius of circle B squared is equal to our change in x. So we could write it as 3 minus 0 or 0 minus 3. But we'll just write it as 3 minus 0. 3 minus 0 squared plus 1 minus 0 squared. Or the radius, or the distance between these two points, is equal to the square root of, let's see, this is 3 squared plus 1 squared.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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So we could write it as 3 minus 0 or 0 minus 3. But we'll just write it as 3 minus 0. 3 minus 0 squared plus 1 minus 0 squared. Or the radius, or the distance between these two points, is equal to the square root of, let's see, this is 3 squared plus 1 squared. This is 9 plus 1. This is the square root of 10. The radius of B is the square root of 10.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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Or the radius, or the distance between these two points, is equal to the square root of, let's see, this is 3 squared plus 1 squared. This is 9 plus 1. This is the square root of 10. The radius of B is the square root of 10. Now they ask us, which of the following points are on circle A, circle B, or both circles? So all we have to do now is look at these points. If this point is a square root of 10 away from point B, then it's on the circle.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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The radius of B is the square root of 10. Now they ask us, which of the following points are on circle A, circle B, or both circles? So all we have to do now is look at these points. If this point is a square root of 10 away from point B, then it's on the circle. It's a radius away. A circle is a locus of all points that are a radius away from the center. If it's 5 square roots of 2 from this point, then it's on circle A.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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If this point is a square root of 10 away from point B, then it's on the circle. It's a radius away. A circle is a locus of all points that are a radius away from the center. If it's 5 square roots of 2 from this point, then it's on circle A. If it's neither, then it's neither. Or it could be both. So let's try these out one by one.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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If it's 5 square roots of 2 from this point, then it's on circle A. If it's neither, then it's neither. Or it could be both. So let's try these out one by one. So point C is at 4, negative 2. So let me color this in a new color. So point C, let me do it in orange.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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So let's try these out one by one. So point C is at 4, negative 2. So let me color this in a new color. So point C, let me do it in orange. Point C is at 1, 2, 3, 4, negative 2. Point C is right over there. Now it looks pretty close.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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So point C, let me do it in orange. Point C is at 1, 2, 3, 4, negative 2. Point C is right over there. Now it looks pretty close. Just this is a hand-drawn drawing, so it's not perfect. So point C is there. It looks pretty close.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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Now it looks pretty close. Just this is a hand-drawn drawing, so it's not perfect. So point C is there. It looks pretty close. But let's actually verify it. The distance between point C and point D, so the distance squared is going to be equal to the change in x's. So we could say 4 minus, so we're trying the distance between C and B.
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Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3
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