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I'm going to draw a circle with my compass. And so let's just pick that point right over there. I could adjust the radius if I like, but I might as well. I'll just leave it right over there. Now, let me draw another circle. And this time, I'm going to center it where the first circle intersects with my line. And then I'm going to adjust the radius to overlap with the first dot, I should say. | Constructing a perpendicular line using a compass and straightedge Geometry Khan Academy.mp3 |
I'll just leave it right over there. Now, let me draw another circle. And this time, I'm going to center it where the first circle intersects with my line. And then I'm going to adjust the radius to overlap with the first dot, I should say. And now, where these two circles intersect, those are points that are equidistant from both of these centers that I just constructed. So let me draw a line that connects those two. And that line is going to be perpendicular to our original line. | Constructing a perpendicular line using a compass and straightedge Geometry Khan Academy.mp3 |
So the first thing that we can think about, these aren't just diagonals. These are lines that are intersecting parallel lines. So you can also view them as transversals. And if we focus on DB right over here, we see that it intersects DC and AB. And since those we know are parallel, this is a parallelogram, we know that alternate interior angles must be congruent. So that angle must be equal to that angle there. And let me make a label here. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
And if we focus on DB right over here, we see that it intersects DC and AB. And since those we know are parallel, this is a parallelogram, we know that alternate interior angles must be congruent. So that angle must be equal to that angle there. And let me make a label here. Let me call that middle point E. So we know that angle ABE must be congruent to angle CDE by alternate interior angles of a transversal intersecting parallel lines. Alternate interior angles. Now if we look at diagonal AC, or we should call it transversal AC, we can make the same argument. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
And let me make a label here. Let me call that middle point E. So we know that angle ABE must be congruent to angle CDE by alternate interior angles of a transversal intersecting parallel lines. Alternate interior angles. Now if we look at diagonal AC, or we should call it transversal AC, we can make the same argument. It intersects here and here. These two lines are parallel. So alternate interior angles must be congruent. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
Now if we look at diagonal AC, or we should call it transversal AC, we can make the same argument. It intersects here and here. These two lines are parallel. So alternate interior angles must be congruent. So angle DEC must be congruent to angle BAE for the exact same reason. Now we have something interesting. If we look at this top triangle over here and this bottom triangle, we have one set of corresponding angles that are congruent. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
So alternate interior angles must be congruent. So angle DEC must be congruent to angle BAE for the exact same reason. Now we have something interesting. If we look at this top triangle over here and this bottom triangle, we have one set of corresponding angles that are congruent. We have a side in between that's going to be congruent. Actually, let me write that down explicitly. We know, and we've proved this to ourselves in the previous video, that parallelograms not only are opposite sides parallel, they are also congruent. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
If we look at this top triangle over here and this bottom triangle, we have one set of corresponding angles that are congruent. We have a side in between that's going to be congruent. Actually, let me write that down explicitly. We know, and we've proved this to ourselves in the previous video, that parallelograms not only are opposite sides parallel, they are also congruent. So we know from the previous video that that side is equal to that side. So let me go back to what I was saying. We have two sets of corresponding angles that are congruent. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
We know, and we've proved this to ourselves in the previous video, that parallelograms not only are opposite sides parallel, they are also congruent. So we know from the previous video that that side is equal to that side. So let me go back to what I was saying. We have two sets of corresponding angles that are congruent. We have a side in between that's congruent. And then we have another set of corresponding angles that are congruent. So we know that this triangle is congruent to that triangle by angle side angle. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
We have two sets of corresponding angles that are congruent. We have a side in between that's congruent. And then we have another set of corresponding angles that are congruent. So we know that this triangle is congruent to that triangle by angle side angle. So we know that triangle, I'm going to go from the blue to the orange to the last one, triangle ABE is congruent to triangle blue, orange, and the last one, CDE, by angle side angle congruency. Now, what does that do for us? Well, we know if two triangles are congruent, all of their corresponding features, especially all of their corresponding sides, are congruent. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
So we know that this triangle is congruent to that triangle by angle side angle. So we know that triangle, I'm going to go from the blue to the orange to the last one, triangle ABE is congruent to triangle blue, orange, and the last one, CDE, by angle side angle congruency. Now, what does that do for us? Well, we know if two triangles are congruent, all of their corresponding features, especially all of their corresponding sides, are congruent. So we know that side EC corresponds to side EA. Or I could say side AE corresponds to side CE. Their corresponding sides of congruent triangles, so their measures or their lengths must be the same. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
Well, we know if two triangles are congruent, all of their corresponding features, especially all of their corresponding sides, are congruent. So we know that side EC corresponds to side EA. Or I could say side AE corresponds to side CE. Their corresponding sides of congruent triangles, so their measures or their lengths must be the same. So AE must be equal to CE. Let me put two slashes since I already used one slash over here. Now, by the same exact logic, we know that DE, we know that, let me focus on this, we know that BE must be equal to DE. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
Their corresponding sides of congruent triangles, so their measures or their lengths must be the same. So AE must be equal to CE. Let me put two slashes since I already used one slash over here. Now, by the same exact logic, we know that DE, we know that, let me focus on this, we know that BE must be equal to DE. Once again, so their corresponding sides of two congruent triangles, so they must have the same length. So this is corresponding sides of congruent triangles. So BE is equal to DE. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
Now, by the same exact logic, we know that DE, we know that, let me focus on this, we know that BE must be equal to DE. Once again, so their corresponding sides of two congruent triangles, so they must have the same length. So this is corresponding sides of congruent triangles. So BE is equal to DE. And we've done our proof. We've shown that, look, diagonal DB is splitting AC into two segments of equal length and vice versa. AC is splitting DB into two segments of equal length. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
So BE is equal to DE. And we've done our proof. We've shown that, look, diagonal DB is splitting AC into two segments of equal length and vice versa. AC is splitting DB into two segments of equal length. So they are bisecting each other. Now let's go the other way around. Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
AC is splitting DB into two segments of equal length. So they are bisecting each other. Now let's go the other way around. Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram. So let me see. So we're going to assume that the two diagonals are bisecting each other. So we're assuming that that is equal to that and that that right over there is equal to that. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram. So let me see. So we're going to assume that the two diagonals are bisecting each other. So we're assuming that that is equal to that and that that right over there is equal to that. Given that, we want to prove that this is a parallelogram. And to do that, we just have to remind ourselves that this angle is going to be equal to that angle. It's one of the first things we learn, because they are vertical angles. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
So we're assuming that that is equal to that and that that right over there is equal to that. Given that, we want to prove that this is a parallelogram. And to do that, we just have to remind ourselves that this angle is going to be equal to that angle. It's one of the first things we learn, because they are vertical angles. So let me write this down. C, let me label this point, CED, angle CED, is going to be equal to or is congruent to angle BEA. And what is that? | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
It's one of the first things we learn, because they are vertical angles. So let me write this down. C, let me label this point, CED, angle CED, is going to be equal to or is congruent to angle BEA. And what is that? Well, that shows us that these two triangles are congruent, because we have a corresponding size that are congruent, an angle in between, and then another side. So we now know that the triangle, I'll keep this in yellow, triangle AEB is congruent to triangle DEC by side angle, side congruency. By SAS, congruent triangles. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
And what is that? Well, that shows us that these two triangles are congruent, because we have a corresponding size that are congruent, an angle in between, and then another side. So we now know that the triangle, I'll keep this in yellow, triangle AEB is congruent to triangle DEC by side angle, side congruency. By SAS, congruent triangles. Fair enough. Now, if we know that two triangles are congruent, we know that all of the corresponding sides and angles are congruent. So for example, we know that angle CDE is going to be congruent to angle BAE. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
By SAS, congruent triangles. Fair enough. Now, if we know that two triangles are congruent, we know that all of the corresponding sides and angles are congruent. So for example, we know that angle CDE is going to be congruent to angle BAE. BAE is going to be congruent to angle BAE, and this is just corresponding angles. And now we have this transversal of these two lines that could be parallel if the alternate interior angles are congruent, and we see that they are. These two are kind of candidate alternate interior angles, and they are congruent. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
So for example, we know that angle CDE is going to be congruent to angle BAE. BAE is going to be congruent to angle BAE, and this is just corresponding angles. And now we have this transversal of these two lines that could be parallel if the alternate interior angles are congruent, and we see that they are. These two are kind of candidate alternate interior angles, and they are congruent. So AB must be parallel to CD. AB is parallel to CD by alternate interior angles congruent of parallel lines. I'm just writing in some shorthand to forgive the cryptic nature of it. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
These two are kind of candidate alternate interior angles, and they are congruent. So AB must be parallel to CD. AB is parallel to CD by alternate interior angles congruent of parallel lines. I'm just writing in some shorthand to forgive the cryptic nature of it. I'm saying it out. And so we can then do the exact same logic. We've just shown that these two sides are parallel. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
I'm just writing in some shorthand to forgive the cryptic nature of it. I'm saying it out. And so we can then do the exact same logic. We've just shown that these two sides are parallel. We can then do the exact same logic to show that these two sides are parallel. And I won't necessarily write it all out, but it's the exact same proof to show that these two. So first of all, we know that this angle is congruent to that angle right over there. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
We've just shown that these two sides are parallel. We can then do the exact same logic to show that these two sides are parallel. And I won't necessarily write it all out, but it's the exact same proof to show that these two. So first of all, we know that this angle is congruent to that angle right over there. And then we know, actually let me write it out. So we know that angle AEC is congruent to angle DEB. They are vertical angles. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
So first of all, we know that this angle is congruent to that angle right over there. And then we know, actually let me write it out. So we know that angle AEC is congruent to angle DEB. They are vertical angles. And then we see that triangle AEC must be congruent to triangle DEB by side angle side. So then we have triangle AEC must be congruent to triangle DEB by SAS congruency. Then we know that corresponding angles must be congruent. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
They are vertical angles. And then we see that triangle AEC must be congruent to triangle DEB by side angle side. So then we have triangle AEC must be congruent to triangle DEB by SAS congruency. Then we know that corresponding angles must be congruent. So that we know that angle, so for example, angle CAE must be congruent to angle BDE. And this is their corresponding angles of congruent triangles. So CAE, let me do this in a new color, must be congruent to BDE. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
Then we know that corresponding angles must be congruent. So that we know that angle, so for example, angle CAE must be congruent to angle BDE. And this is their corresponding angles of congruent triangles. So CAE, let me do this in a new color, must be congruent to BDE. And now we have a transversal. So the alternate interior angles are congruent. So the two lines that the transversal is intersecting must be parallel. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
So CAE, let me do this in a new color, must be congruent to BDE. And now we have a transversal. So the alternate interior angles are congruent. So the two lines that the transversal is intersecting must be parallel. So this must be parallel to that. So then we have AC must be parallel to BD by alternate interior angles. And we're done. | Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3 |
So in this first triangle right over here, we're given that this side has length 3, this side has length 6. And this little dotted line here, this is clearly the angle bisector, because they're telling us that this angle is congruent to that angle right over there. And then they tell us that the length of just this part of this side right over here is 2. So from here to here is 2. And that this length is x. So let's figure out what x is. So the angle bisector theorem tells us that the ratio of 3 to 2 is going to be equal to 6 to x. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
So from here to here is 2. And that this length is x. So let's figure out what x is. So the angle bisector theorem tells us that the ratio of 3 to 2 is going to be equal to 6 to x. And we can just solve for x. So 3 to 2 is going to be equal to 6 to x. And then once again, you can just cross multiply, or you can multiply both sides by 2 and x. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
So the angle bisector theorem tells us that the ratio of 3 to 2 is going to be equal to 6 to x. And we can just solve for x. So 3 to 2 is going to be equal to 6 to x. And then once again, you can just cross multiply, or you can multiply both sides by 2 and x. It kind of gives you the same result. If you cross multiply, you get 3x. 3x is equal to 2 times 6 is 12. x is equal to, divide both sides by 3. x is equal to 4. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
And then once again, you can just cross multiply, or you can multiply both sides by 2 and x. It kind of gives you the same result. If you cross multiply, you get 3x. 3x is equal to 2 times 6 is 12. x is equal to, divide both sides by 3. x is equal to 4. So in this case, x is equal to 4. And this is kind of interesting, because we just realize now that this side, this entire side right over here, is going to be equal to 6. So even though it doesn't look that way based on how it's drawn, this is actually an isosceles triangle. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
3x is equal to 2 times 6 is 12. x is equal to, divide both sides by 3. x is equal to 4. So in this case, x is equal to 4. And this is kind of interesting, because we just realize now that this side, this entire side right over here, is going to be equal to 6. So even though it doesn't look that way based on how it's drawn, this is actually an isosceles triangle. It has a 6 and a 6. And then the base right over here is 3. It's kind of interesting. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
So even though it doesn't look that way based on how it's drawn, this is actually an isosceles triangle. It has a 6 and a 6. And then the base right over here is 3. It's kind of interesting. Over here, we're given that this length is 5. This length is 7. This entire side is 10. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
It's kind of interesting. Over here, we're given that this length is 5. This length is 7. This entire side is 10. And then we have this angle bisector right over there. And we need to figure out just this part of the triangle between this point, if we call this point A, and this point right over here. We need to find the length of AB right over here. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
This entire side is 10. And then we have this angle bisector right over there. And we need to figure out just this part of the triangle between this point, if we call this point A, and this point right over here. We need to find the length of AB right over here. So once again, angle bisector theorem, the ratio of 5 to this. Let me do this in a new color. The ratio of 5 to x is going to be equal to the ratio of 7 to this distance right over here. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
We need to find the length of AB right over here. So once again, angle bisector theorem, the ratio of 5 to this. Let me do this in a new color. The ratio of 5 to x is going to be equal to the ratio of 7 to this distance right over here. And what is that distance? Well, if the whole thing is 10, and this is x, then this distance right over here is going to be 10 minus x. So the ratio of 5 to x is equal to 7 over 10 minus x. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
The ratio of 5 to x is going to be equal to the ratio of 7 to this distance right over here. And what is that distance? Well, if the whole thing is 10, and this is x, then this distance right over here is going to be 10 minus x. So the ratio of 5 to x is equal to 7 over 10 minus x. And we can cross multiply. 5 times 10 minus x is 50 minus 5x. And then x times 7 is equal to 7x. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
So the ratio of 5 to x is equal to 7 over 10 minus x. And we can cross multiply. 5 times 10 minus x is 50 minus 5x. And then x times 7 is equal to 7x. Add 5x to both sides of this equation. You get 50 is equal to 12x. We can divide both sides by 12. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
And then x times 7 is equal to 7x. Add 5x to both sides of this equation. You get 50 is equal to 12x. We can divide both sides by 12. And we get 50 over 12 is equal to x. And we can reduce this. Let's see, if you divide the numerator and the denominator by 2, you get this is the same thing as 25 over 6, which is the same thing, if we want to write it as a mixed number, as 4. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
We can divide both sides by 12. And we get 50 over 12 is equal to x. And we can reduce this. Let's see, if you divide the numerator and the denominator by 2, you get this is the same thing as 25 over 6, which is the same thing, if we want to write it as a mixed number, as 4. 24 over 6 is 4. And then you have 1 6 left over. 4 and 1 6. | Angle bisector theorem examples Geometry Khan Academy.mp3 |
Let's see, if you divide the numerator and the denominator by 2, you get this is the same thing as 25 over 6, which is the same thing, if we want to write it as a mixed number, as 4. 24 over 6 is 4. And then you have 1 6 left over. 4 and 1 6. So this length right over here is going to, oh sorry, this length right over here, x is 4 and 1 6. And then this length over here is going to be 10 minus 4 and 1 6. So what is that? | Angle bisector theorem examples Geometry Khan Academy.mp3 |
And what we need to do, we need to prove that the area of this rhombus is equal to 1 half times AC times BD. So we're essentially proving that the area of a rhombus is 1 half times the product of the lengths of its diagonals. So let's see what we can do over here. So there's a bunch of things we know about rhombi, and all rhombi are parallelograms, so there's tons of things that we know about parallelograms. First of all, if it's a rhombus, we know that all of the sides are congruent. So that side length is equal to that side length is equal to that side length is equal to that side length. Because it's a parallelogram, we know that diagonals bisect each other, so we know that this length, let me call this point over here B, let's call this E, we know that BE is going to be equal to ED. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
So there's a bunch of things we know about rhombi, and all rhombi are parallelograms, so there's tons of things that we know about parallelograms. First of all, if it's a rhombus, we know that all of the sides are congruent. So that side length is equal to that side length is equal to that side length is equal to that side length. Because it's a parallelogram, we know that diagonals bisect each other, so we know that this length, let me call this point over here B, let's call this E, we know that BE is going to be equal to ED. So that's BE, we know that's going to be equal to ED. And we know that AE is equal to EC. We also know because this is a rhombus, and we proved this in the last video, that the diagonals, not only do they bisect each other, but they are also perpendicular. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
Because it's a parallelogram, we know that diagonals bisect each other, so we know that this length, let me call this point over here B, let's call this E, we know that BE is going to be equal to ED. So that's BE, we know that's going to be equal to ED. And we know that AE is equal to EC. We also know because this is a rhombus, and we proved this in the last video, that the diagonals, not only do they bisect each other, but they are also perpendicular. So we know that this is a right angle, this is a right angle, that is a right angle, and then this is a right angle. So the easiest way to think about it is, if we can show that this triangle ADC is congruent to triangle ABC, and if we can figure out the area of one of them, we can just double it. So the first part is pretty straightforward. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
We also know because this is a rhombus, and we proved this in the last video, that the diagonals, not only do they bisect each other, but they are also perpendicular. So we know that this is a right angle, this is a right angle, that is a right angle, and then this is a right angle. So the easiest way to think about it is, if we can show that this triangle ADC is congruent to triangle ABC, and if we can figure out the area of one of them, we can just double it. So the first part is pretty straightforward. So we can see that triangle ADC, we know that triangle ADC is going to be congruent to triangle ABC, and we know that by side-side-side congruency. This side is congruent to that side, this side is congruent to that side, and they both share AC right over here. So this is by side-side-side. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
So the first part is pretty straightforward. So we can see that triangle ADC, we know that triangle ADC is going to be congruent to triangle ABC, and we know that by side-side-side congruency. This side is congruent to that side, this side is congruent to that side, and they both share AC right over here. So this is by side-side-side. And so we can say that the area, so because of that, we know that the area of ABCD is just going to be equal to 2 times the area of, we can pick either one of these, we can say 2 times the area of ABC. Because the area of ABCD, actually let me write it this way. The area of ABCD is equal to the area of ADC plus the area of ABC. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
So this is by side-side-side. And so we can say that the area, so because of that, we know that the area of ABCD is just going to be equal to 2 times the area of, we can pick either one of these, we can say 2 times the area of ABC. Because the area of ABCD, actually let me write it this way. The area of ABCD is equal to the area of ADC plus the area of ABC. But since they're congruent, these two are going to be the same thing, so it's just going to be 2 times the area of ABC. Now what is the area of ABC? Well, area of a triangle is just 1 half base times height. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
The area of ABCD is equal to the area of ADC plus the area of ABC. But since they're congruent, these two are going to be the same thing, so it's just going to be 2 times the area of ABC. Now what is the area of ABC? Well, area of a triangle is just 1 half base times height. So area of ABC is just equal to 1 half times the base of that triangle times its height, which is equal to 1 half. What is the length of the base? Well, the length of the base is AC, so it's 1 half, I'll color code it. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
Well, area of a triangle is just 1 half base times height. So area of ABC is just equal to 1 half times the base of that triangle times its height, which is equal to 1 half. What is the length of the base? Well, the length of the base is AC, so it's 1 half, I'll color code it. The base is AC, and then what is the height? What is the height here? Well, we know that this diagonal right over here, that's a perpendicular bisector. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
Well, the length of the base is AC, so it's 1 half, I'll color code it. The base is AC, and then what is the height? What is the height here? Well, we know that this diagonal right over here, that's a perpendicular bisector. So the height is just the distance from BE. So it's AC times BE. That is the height. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
Well, we know that this diagonal right over here, that's a perpendicular bisector. So the height is just the distance from BE. So it's AC times BE. That is the height. This is an altitude. It intersects this base at a 90 degree angle. Or we could say BE is the same thing as 1 half times BD. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
That is the height. This is an altitude. It intersects this base at a 90 degree angle. Or we could say BE is the same thing as 1 half times BD. So this is equal to 1 half times AC. That's our base. And then our height is BE, which we're saying is the same thing as 1 half times BD. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
Or we could say BE is the same thing as 1 half times BD. So this is equal to 1 half times AC. That's our base. And then our height is BE, which we're saying is the same thing as 1 half times BD. Which is 1 half times BD. So that's the area of just ABC. That's just the area of this broader triangle right up there, or that larger triangle right up there, that half of the rhombus. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
And then our height is BE, which we're saying is the same thing as 1 half times BD. Which is 1 half times BD. So that's the area of just ABC. That's just the area of this broader triangle right up there, or that larger triangle right up there, that half of the rhombus. But we just said that the area of the whole thing is 2 times that. So if we go back, if we use both this information and this information right over here, we have the area of ABCD is going to be equal to 2 times the area of ABC. But the area of ABC is this thing right over here. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
That's just the area of this broader triangle right up there, or that larger triangle right up there, that half of the rhombus. But we just said that the area of the whole thing is 2 times that. So if we go back, if we use both this information and this information right over here, we have the area of ABCD is going to be equal to 2 times the area of ABC. But the area of ABC is this thing right over here. It is. So 2 times the area of ABC, area of ABC is that right over there. So 1 half times 1 half is 1 fourth times AC times BD. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
But the area of ABC is this thing right over here. It is. So 2 times the area of ABC, area of ABC is that right over there. So 1 half times 1 half is 1 fourth times AC times BD. And then you see where this is going. 2 times 1 fourth is 1 half times AC times BD. Fairly straightforward. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
So 1 half times 1 half is 1 fourth times AC times BD. And then you see where this is going. 2 times 1 fourth is 1 half times AC times BD. Fairly straightforward. Which is a neat result. And actually I haven't done this in a video yet. I'll do it in the next video. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
Fairly straightforward. Which is a neat result. And actually I haven't done this in a video yet. I'll do it in the next video. There are other ways of finding the areas of parallelograms generally. It's essentially base times height. But for a rhombus, we could do that because it is a parallelogram. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
I'll do it in the next video. There are other ways of finding the areas of parallelograms generally. It's essentially base times height. But for a rhombus, we could do that because it is a parallelogram. But we also have this other neat little result that we proved in this video. That if we know the lengths of the diagonals, the area of the rhombus is 1 half times the products of the lengths of the diagonals. Which is kind of a neat result. | Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3 |
Let's just do a ton of more examples just so that we make sure that we're getting this trig function thing down well. So let's construct ourselves some right triangles. Let's construct ourselves some right triangles. And I want to be very clear, the way I've defined it so far, this will only work in right triangles. So if you're trying to find the trig functions of angles that aren't part of right triangles, we're going to see that we're going to have to construct right triangles. But let's just focus on the right triangles for now. So let's say that I have a triangle where let's say this length down here is 7. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
And I want to be very clear, the way I've defined it so far, this will only work in right triangles. So if you're trying to find the trig functions of angles that aren't part of right triangles, we're going to see that we're going to have to construct right triangles. But let's just focus on the right triangles for now. So let's say that I have a triangle where let's say this length down here is 7. And let's say the length of this side up here, let's say that that is 4. And let's figure out what the hypotenuse over here is going to be. So we know, let's call the hypotenuse h. We know that h squared is going to be equal to 7 squared plus 4 squared. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
So let's say that I have a triangle where let's say this length down here is 7. And let's say the length of this side up here, let's say that that is 4. And let's figure out what the hypotenuse over here is going to be. So we know, let's call the hypotenuse h. We know that h squared is going to be equal to 7 squared plus 4 squared. We know that from the Pythagorean theorem. That the hypotenuse squared is equal to the sum of the squares of the other two sides. h squared is equal to 7 squared plus 4 squared. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
So we know, let's call the hypotenuse h. We know that h squared is going to be equal to 7 squared plus 4 squared. We know that from the Pythagorean theorem. That the hypotenuse squared is equal to the sum of the squares of the other two sides. h squared is equal to 7 squared plus 4 squared. So this is equal to 49 plus 16. 49 plus 10 is 59. Plus 6 is 65. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
h squared is equal to 7 squared plus 4 squared. So this is equal to 49 plus 16. 49 plus 10 is 59. Plus 6 is 65. So this h squared, let me write h squared, it's a different shade of yellow. So we have h squared is equal to 65. Did I do that right? | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Plus 6 is 65. So this h squared, let me write h squared, it's a different shade of yellow. So we have h squared is equal to 65. Did I do that right? 49 plus 10 is 59. Plus another 6 is 65. Or we could say that h is equal to, if we take the square root of both sides, square root of 65. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Did I do that right? 49 plus 10 is 59. Plus another 6 is 65. Or we could say that h is equal to, if we take the square root of both sides, square root of 65. And we really can't simplify this at all. This is 13, this is the same thing as 13 times 5. Both of those are not perfect squares and they're both primes. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Or we could say that h is equal to, if we take the square root of both sides, square root of 65. And we really can't simplify this at all. This is 13, this is the same thing as 13 times 5. Both of those are not perfect squares and they're both primes. You can't simplify this anymore. So this is equal to the square root of 65. Now let's find the trig functions for this angle up here. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Both of those are not perfect squares and they're both primes. You can't simplify this anymore. So this is equal to the square root of 65. Now let's find the trig functions for this angle up here. Let's call that angle up there theta. So whenever you do it, you always want to write down, at least for me it works out to write down, SOH CAH TOA. I have these vague memories of my trigonometry teacher. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Now let's find the trig functions for this angle up here. Let's call that angle up there theta. So whenever you do it, you always want to write down, at least for me it works out to write down, SOH CAH TOA. I have these vague memories of my trigonometry teacher. Maybe I read it in some book. I don't know about some type of Indian princess named SOH CAH TOA or whatever. But it's a very useful mnemonic. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
I have these vague memories of my trigonometry teacher. Maybe I read it in some book. I don't know about some type of Indian princess named SOH CAH TOA or whatever. But it's a very useful mnemonic. So we can apply SOH CAH TOA. Let's say we wanted to find the cosine. We want to find the cosine of our angle. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
But it's a very useful mnemonic. So we can apply SOH CAH TOA. Let's say we wanted to find the cosine. We want to find the cosine of our angle. You say SOH CAH TOA. So the CAH tells us what to do with cosine. The CAH part tells us that cosine is adjacent over hypotenuse. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
We want to find the cosine of our angle. You say SOH CAH TOA. So the CAH tells us what to do with cosine. The CAH part tells us that cosine is adjacent over hypotenuse. Cosine is equal to adjacent over hypotenuse. So let's look over here. To theta, what side is adjacent? | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
The CAH part tells us that cosine is adjacent over hypotenuse. Cosine is equal to adjacent over hypotenuse. So let's look over here. To theta, what side is adjacent? Well, we know that the hypotenuse is this side over here. So it can't be that side. The only other side that's kind of adjacent to it, that isn't the hypotenuse, is this 4. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
To theta, what side is adjacent? Well, we know that the hypotenuse is this side over here. So it can't be that side. The only other side that's kind of adjacent to it, that isn't the hypotenuse, is this 4. So the adjacent side over here, that side is literally right next to the angle. It's one of the sides that kind of forms the angle. It's 4 over the hypotenuse. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
The only other side that's kind of adjacent to it, that isn't the hypotenuse, is this 4. So the adjacent side over here, that side is literally right next to the angle. It's one of the sides that kind of forms the angle. It's 4 over the hypotenuse. The hypotenuse, we already know, is square root of 65. So it's 4 over the square root of 65. And sometimes people want you to rationalize the denominator, which means they don't like to have an irrational number in the denominator, like the square root of 65. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
It's 4 over the hypotenuse. The hypotenuse, we already know, is square root of 65. So it's 4 over the square root of 65. And sometimes people want you to rationalize the denominator, which means they don't like to have an irrational number in the denominator, like the square root of 65. And if you want to rewrite this without an irrational number in the denominator, you can multiply the numerator and the denominator by square root of 65. This clearly will not change the number because we're multiplying it by something over itself. So we're multiplying the number by 1. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
And sometimes people want you to rationalize the denominator, which means they don't like to have an irrational number in the denominator, like the square root of 65. And if you want to rewrite this without an irrational number in the denominator, you can multiply the numerator and the denominator by square root of 65. This clearly will not change the number because we're multiplying it by something over itself. So we're multiplying the number by 1. That won't change the number, but at least it gets rid of the irrational number in the denominator. So the numerator becomes 4 times the square root of 65. And the denominator, square root of 65 times square root of 65, is just going to be 65. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
So we're multiplying the number by 1. That won't change the number, but at least it gets rid of the irrational number in the denominator. So the numerator becomes 4 times the square root of 65. And the denominator, square root of 65 times square root of 65, is just going to be 65. We didn't get rid of the irrational number. It's still there, but it's now in the numerator. Now let's do the other trig functions, or at least the other core trig functions. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
And the denominator, square root of 65 times square root of 65, is just going to be 65. We didn't get rid of the irrational number. It's still there, but it's now in the numerator. Now let's do the other trig functions, or at least the other core trig functions. We'll learn in the future that there's actually a ton of them, but they're all derived from these. So let's think about what the sine of theta is. Once again, go to SOH CAH TOA. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Now let's do the other trig functions, or at least the other core trig functions. We'll learn in the future that there's actually a ton of them, but they're all derived from these. So let's think about what the sine of theta is. Once again, go to SOH CAH TOA. The SOH tells us what to do with sine. Sine is opposite over hypotenuse. So for this angle, what side is opposite? | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Once again, go to SOH CAH TOA. The SOH tells us what to do with sine. Sine is opposite over hypotenuse. So for this angle, what side is opposite? Well, you just go opposite it. What it opens into, it's opposite the 7. So the opposite side is the 7. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
So for this angle, what side is opposite? Well, you just go opposite it. What it opens into, it's opposite the 7. So the opposite side is the 7. This is right here. That is the opposite side. And then the hypotenuse, it's opposite over hypotenuse. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
So the opposite side is the 7. This is right here. That is the opposite side. And then the hypotenuse, it's opposite over hypotenuse. The hypotenuse is the square root of 65. And once again, if we wanted to rationalize this, we could multiply it times the square root of 65 over the square root of 65. In the numerator, we'll get 7 square roots of 65. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
And then the hypotenuse, it's opposite over hypotenuse. The hypotenuse is the square root of 65. And once again, if we wanted to rationalize this, we could multiply it times the square root of 65 over the square root of 65. In the numerator, we'll get 7 square roots of 65. And in the denominator, we will get just 65 again. Now let's do tangent. So if I asked you the tangent of theta, once again, go back to SOH CAH TOA. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
In the numerator, we'll get 7 square roots of 65. And in the denominator, we will get just 65 again. Now let's do tangent. So if I asked you the tangent of theta, once again, go back to SOH CAH TOA. The TOA part tells us what to do with tangent. It tells us that tangent is equal to opposite over adjacent. So for this angle, what is opposite? | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
So if I asked you the tangent of theta, once again, go back to SOH CAH TOA. The TOA part tells us what to do with tangent. It tells us that tangent is equal to opposite over adjacent. So for this angle, what is opposite? We've already figured it out. It's 7. It opens into the 7. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
So for this angle, what is opposite? We've already figured it out. It's 7. It opens into the 7. It's opposite the 7. So it's 7 over what side is adjacent? Well, this 4 is adjacent. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
It opens into the 7. It's opposite the 7. So it's 7 over what side is adjacent? Well, this 4 is adjacent. So the adjacent side is 4. So it's 7 over 4. And we're done. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Well, this 4 is adjacent. So the adjacent side is 4. So it's 7 over 4. And we're done. We figured out all of the trig ratios for theta. Let's do another one. And I'll make it a little bit concrete. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
And we're done. We figured out all of the trig ratios for theta. Let's do another one. And I'll make it a little bit concrete. Because right now we've been saying, oh, what's tangent of x? What's tangent of theta? Let's make it a little bit more concrete. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
And I'll make it a little bit concrete. Because right now we've been saying, oh, what's tangent of x? What's tangent of theta? Let's make it a little bit more concrete. Let's say, let me draw another right triangle. Everything we're dealing with, these are going to be right triangles. Let's say the hypotenuse has length 4. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Let's make it a little bit more concrete. Let's say, let me draw another right triangle. Everything we're dealing with, these are going to be right triangles. Let's say the hypotenuse has length 4. Let's say that this side over here has length 2. And let's say that this length over here is going to be 2 times the square root of 3. We can verify that this works. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Let's say the hypotenuse has length 4. Let's say that this side over here has length 2. And let's say that this length over here is going to be 2 times the square root of 3. We can verify that this works. If you have this side squared, so you have, let me write it down, 2 times the square root of 3 squared plus 2 squared is equal to what? This is 2. This is going to be 4 times 3. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
We can verify that this works. If you have this side squared, so you have, let me write it down, 2 times the square root of 3 squared plus 2 squared is equal to what? This is 2. This is going to be 4 times 3. 4 times 3 plus 4. And this is going to be equal to 12 plus 4 is equal to 16. And 16 is indeed 4 squared. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
This is going to be 4 times 3. 4 times 3 plus 4. And this is going to be equal to 12 plus 4 is equal to 16. And 16 is indeed 4 squared. So this does equal 4 squared. It does equal 4 squared. It satisfies the Pythagorean theorem. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
And 16 is indeed 4 squared. So this does equal 4 squared. It does equal 4 squared. It satisfies the Pythagorean theorem. And if you remember some of your work from 30-60-90 triangles that you might have learned in geometry, you might recognize that this is a 30-60-90 triangle. This right here is our right angle. I should have drawn it from the get-go to show that this is a right triangle. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
It satisfies the Pythagorean theorem. And if you remember some of your work from 30-60-90 triangles that you might have learned in geometry, you might recognize that this is a 30-60-90 triangle. This right here is our right angle. I should have drawn it from the get-go to show that this is a right triangle. This angle right over here is our 30-degree angle. And then this angle up here is a 60-degree angle. And it's a 30-60-90 because the side opposite the 30 degrees is half the hypotenuse. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
I should have drawn it from the get-go to show that this is a right triangle. This angle right over here is our 30-degree angle. And then this angle up here is a 60-degree angle. And it's a 30-60-90 because the side opposite the 30 degrees is half the hypotenuse. And then the side opposite the 60 degrees is the square root of 3 times the other side that's not the hypotenuse. So with that said, this isn't supposed to be a review of 30-60-90 triangles, although I just did it. Let's actually find the trig ratios for the different angles. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
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