2005 AMC 8 Problems/Problem 25

Problem 25

A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?

[asy] pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(a--d--b--c--cycle); draw(circle(o, 2.5)); [/asy]

$\textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}$


Solution 1

Let the region within the circle and square be $a$. In other words, it is the area inside the circle $\textbf{and}$ the square. Let $r$ be the radius. We know that the area of the circle minus $a$ is equal to the area of the square, minus $a$ .

We get:

$\pi r^2 -a=4-a$



So the answer is $\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}$.

Solution 2

We realize that since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square.

$\pi r^2=4$



So the answer is $\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}$.

Video Solution

https://youtu.be/dWxzg4zmGOs Soo, DRMS, NM

Video Solution by OmegaLearn


Video Solution

https://www.youtube.com/watch?v=xdeWCR666q8 ~David

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
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