# Math Help - AMC problem, finding the area inside a circle but outside a square

1. ## AMC problem, finding the area inside a circle but outside a square

Hi everyone. I'm a little stuck on how to solve this problem.

2010 AMC 10B Problems - AoPSWiki

Basically, there's a circle and a square with the same center. I can't just take the area of the circle minus the area of the square, since some of the square lies outside of the circle. How do I take that into account?

The answer, by the way, should be "B."

2. Originally Posted by Lord Darkin
Hi everyone. I'm a little stuck on how to solve this problem.

2010 AMC 10B Problems - AoPSWiki

Basically, there's a circle and a square with the same center. I can't just take the area of the circle minus the area of the square, since some of the square lies outside of the circle. How do I take that into account?

The answer, by the way, should be "B."

-Dan

3. Yes, that's the problem (16) and thanks for the link.

So I played around using those formulas, and also analyzed the circle/square situation further. Apparently the triangle that is below the "circular segment" as depicted by wikipedia is an equilateral triangle, with sides of root3 over 3. Therefore theta = 60 degrees.

By using the formula of wikipedia, and multiplying it by 4 to take into account the 4 circular regions out of the square, I ended up with

$
A = \frac{\pi}{3} - 4(\frac{1}{6}(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}))
$

Where A should be the answer to the question. But I get...

$A = \frac{-\pi}{9} + \frac{\sqrt{3}}{3}$

Something isn't right here ....

Edit: I should have said this before, but pi/3 for the first term represents the area of the entire circle.

4. Area of a sector = $\pi\left(\dfrac{\sqrt3}{3}\right)^2\cdot \dfrac{60}{360} = \dfrac{\pi}{18}$

Area of the triangle = $\dfrac12 ab \sin c = \dfrac12 \left(\dfrac{\sqrt3}{3}\right)^2 \sin (60) = \dfrac{\sqrt3}{12}$

Area required = $4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right)$

5. Thanks, I forgot sin60 = 1/sqrt3