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• 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

• 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 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.

The center of this circle is O. And I

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B1 INT squared equal oa circle hypotenuse pythagorean theorem

# Pythagorean theorem and radii of circles | Circles | Geometry | Khan Academy

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onyi posted on 2016/02/21
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