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# In the circle above, if the area of the rectangle set inside

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In the circle above, if the area of the rectangle set inside [#permalink]  23 Mar 2018, 03:57
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In the circle above, if the area of the rectangle set inside the circle is 200 and b = 8a, what is the circumference of the circle?

A. $$25\pi \sqrt{65}$$

B. $$5 \sqrt{65}$$

C. $$5\pi \sqrt{13}$$

D. $$5\pi \sqrt{65}$$

E. $$5\pi$$
[Reveal] Spoiler: OA

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Re: In the circle above, if the area of the rectangle set inside [#permalink]  27 Mar 2018, 19:19
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If you don't know the radius of a circle, you can't really answer anything about the circle, so finding the radius is your first priority. In this case, the diameter of the circle is the diagonal of this rectangle, so we should start there.

Since one side of the circle is a and the other is b, but b equals 8a, we can say the two sides are a and 8a. If the area is 200, we know that 8a^2 = 200, so a must equal 5. Therefore, the two sides of the rectangle are 5 and (plugging 5 into 8a gives us) 40 for the other side.

If two sides of a right triangle are 5 and 40, this isn't a special triangle, so we're stuck with the pythagorean theorem:

5^2 + 40^2 = d^2 (I'm calling the hypotenuse d here.)

So:

25 + 1600 = d^2

and

1625 = d^2

so

d = √1625

Now what? Nobody expects you to know the square root of 1625 but if you start factoring it out, we should see an answer pretty quickly. 1625 is obviously divisible by 5. An easy way to divide anything by 5 is to instead divide by 10 and then double it. In this case, dividing 1625 by 10 gets us 162.5, and doubling that gets us 325. Dividing 325 by 10 again gets us 32.5, and doubling that we get 65. So 1625 = 5x5x65. We know that 65 is 13x5, but since there's no nice square root of 65 we'll just leave it alone for now.

So we have:

d = √(5x5x65) = 5√65

This is the diameter of the circle, and since the circumference of a circle is πd, the circumference of this circle must be 5π√65, giving us answer choice D.
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Re: In the circle above, if the area of the rectangle set inside   [#permalink] 27 Mar 2018, 19:19
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