AMC10 2006 10B, Problem 8

A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle? [asy] defaultpen(linewidth(0.8)); size(100); real r=sqrt(50), s=sqrt(10); draw(Arc(origin, r, 0, 180)); draw((r,0)--(-r,0), dashed); draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle); [/asy]

Solution: Since the area of the square is $40$, the length of a side is $\sqrt{40}=2\sqrt{10}$. The distance between the center of the semicircle and one of the bottom vertices of the square is half the length of the side, which is $\sqrt{10}$. Using the Pythagorean Theorem to find the radius $r$ of the semicircle, $r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50$. So, the area of the semicircle is $\frac{1}{2}\cdot \pi \cdot 50 = \fbox{25\pi}$.

Answer: 25\pi