2006 AMC 10B Problem 8

Let the side of the square be $s$, so $s^2 = 40$. The square's base sits on the diameter and, by symmetry, is centered at the semicircle's center $O$. So $O$ is $\frac{s}{2}$ from each bottom corner.

Draw the segment from $O$ to an upper corner of the square; that corner is on the arc, so the segment is a radius $r$. It is the hypotenuse of a right triangle with horizontal leg $\frac{s}{2}$ and vertical leg $s$:
$$r^2 = s^2 + \left(\frac{s}{2}\right)^2 = \frac{5}{4}s^2 = \frac{5}{4} \cdot 40 = 50.$$

The semicircle's area is $\frac{1}{2}\pi r^2 = \frac{1}{2} \cdot 50\pi = 25\pi$.

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

"Inscribed" means the square's vertices touch the arc, so a radius drawn to a vertex creates a right triangle with the square's sides. Working with $s^2$ throughout, never $s$ itself, keeps the square root of $40$ out of the computation entirely.

• Taking the square's side as the radius (r^2 = 40) or as the diameter, giving 20 pi or 5 pi.
• Reporting the full circle's area $50\pi$ (choice (E)) instead of the semicircle's.