2006 AMC 10B Problem 8

1. Let the square's side be $s$. Its area is $40$, so $s^2 = 40$. We will not need $s$ itself, only $s^2$.
2. The square sits on the flat edge of the semicircle, right in the middle. So the center $O$ of the semicircle is at the midpoint of the square's bottom side, and each bottom corner is $\frac{s}{2}$ away from $O$.
3. The top corners of the square touch the curved arc. So the segment from $O$ to a top corner is a radius $r$.
4. That segment is the long side of a right triangle. Go from $O$ sideways $\frac{s}{2}$ to a bottom corner, then straight up $s$ to the top corner. By the Pythagorean theorem,
$$r^2 = \left(\frac{s}{2}\right)^2 + s^2 = \frac{s^2}{4} + s^2 = \frac{40}{4} + 40 = 10 + 40 = 50.$$
5. A semicircle is half a circle, so its area is
$$\frac{1}{2}\pi r^2 = \frac{1}{2} \cdot 50\pi = 25\pi.$$

The answer is $\boxed{\textbf{(B)}\ 25\pi}$.

When a shape is drawn inside a circle with its corners on the curve, a line from the center to a corner is a radius. That line usually forms a right triangle with the shape's sides, so the Pythagorean theorem gives the radius. Here it helps to keep working with $s^2 = 40$ instead of finding $s = \sqrt{40}$; the square roots never appear.