字幕表 動画を再生する 英語字幕をプリント The center of this circle is O. And I apologize if I'm a little out of breath, I actually just did some pull-ups. Anyway, the center of this circle is O. Find the exact length of OA, CD, and OF. So let's look at each of these. So CD, it's part of a right triangle. It's one of the two shorter sides of a right triangle. But we don't know the hypotenuse, so we're not going to be able to figure out CD out right, just yet. Same thing with OF. OF is one of the two shorter sides of a right triangle. But we don't know its hypotenuse either. Now let's look at OA. OA is the hypotenuse of a right triangle, and they've given us the two other sides. So we can use the Pythagorean theorem to figure out the hypotenuse. So we know that 7 squared, let's call this x. We know that 7 squared plus 24 squared is going to be equal to the length of OA squared. It's going to be equal to x squared. 7 squared is 49, and 24 squared, well let's do a multiplication right over here to figure out 24 times 24. 4 times 4 is 16. 4 times 2 is 8 plus-- so that's 96. And then two times 24 is 48. Add them together, we get 6. 9 plus 8 is 17, 576. So 49 plus 576 is equal to x squared. And so let's think about what this is going to be. And this is going to be the same thing as 50 plus 575. I just took one away from this and added one here. So 50 plus 575 is 625. So 625 is equal to x squared. And you might recognize that 25 times 25 is 625. So x is equal to 25. And if you don't believe me you could multiply that out on your own. So x is equal to 25. Or another way of thinking about it, the exact length of OA is equal to 25. Now, how can we somehow use that information to figure out this other stuff? Well all of these other right triangles, all of their hypotenuses are a radius of the circle, and so is OA. OA is a radius of the circle. OG is a radius of the circle. OC is a radius of the circle. Well, we just figured out the radius of the circle is 25. So OG is going to be 25, and OC is going to be 25 as well. So now we just have to apply the Pythagorean theorem a few more times. So right over here, if I call OF, let's just call that, I don't know for the sake of argument, let's call that length equal to a. So here, for this triangle, we see that a squared plus the square root of 141 squared-- I'll just write that as 141-- so plus 141 is going to be equal to 25 squared, which we already know to be 625. If we subtract 141 from both sides let's see where do we get. So let's do 625 minus 141 we get 5 minus 1 is 4. And then 12 here, and we can put a 5 there. 12 minus 4 is 8. 5 minus 1 is 4. So we get 484. So we get a squared is equal to 484. So what squared is equal to 484? Actually, I'll just try to do a prime factorization here to figure this out. So 484 is 2 times 242, which is 2 times 121, which is the same thing as 11 times 11. So another way of thinking about it is this is 2 squared-- so 484, I'll write it over here. 484 is equal to 2 squared times 11 squared. Or it's the same thing as 2 times 11 squared, which is 22 squared. So in this case a is equal to-- let me just clean all that up so I have some space to work with-- a is equal to 22. Let me write that down. a is equal to 22, so that's equal to 22 right here. And that's the length of segment OF. So this is 22. And then finally CD, once again we just apply the Pythagorean theorem. Let's just call this, I don't know, I've already used a. I've already used x. I don't know, I'll call this b. So we see that b squared plus 15 squared, which is the same thing as 225-- 15 squared is 225-- is going to be equal to 25 squared, is going to be equal to 625. Subtract 225 from both sides you get b squared is equal to 400, and the square root of 400 is pretty easy to calculate. B is equal to 20. So segment OA is 25, CD is 20, and OF is 22.
B1 中級 ピタゴラスの定理と円の半径|円|幾何学|カーンアカデミー (Pythagorean theorem and radii of circles | Circles | Geometry | Khan Academy) 64 1 onyi に公開 2021 年 01 月 14 日 シェア シェア 保存 報告 動画の中の単語