2025 AMC 10A Problem 10

Let the large semicircle have radius $R$ and centre $O$ on $\overline{AB}$, and let the small semicircle have radius $r$. Let $h$ be the distance from $\overline{AB}$ to the parallel chord $\overline{CD}$.

The small semicircle has its diameter on $\overline{AB}$, so its highest point is exactly $r$ above $\overline{AB}$. Being tangent to $\overline{CD}$ means that highest point lies on $\overline{CD}$, so
$$r = h.$$

Now drop the perpendicular from $O$ to $\overline{CD}$. It bisects the chord, so it meets $\overline{CD}$ at distance $8$ from $C$, and the radius $OC = R$ is the hypotenuse of a right triangle with legs $8$ and $h$:
$$R^2 = 8^2 + h^2 = 64 + r^2, \qquad\text{so}\qquad R^2 - r^2 = 64.$$

The shaded figure is the large semicircle with the small one removed:
$$\frac{1}{2}\pi R^2 - \frac{1}{2}\pi r^2 = \frac{\pi}{2}\left(R^2 - r^2\right) = \frac{\pi}{2}\cdot 64 = 32\pi.$$

The answer is $\boxed{\textbf{(C)}\ 32\pi}$.

Neither $R$ nor $r$ is determined by the given data, and neither is the position of the small semicircle along $\overline{AB}$; only the combination $R^2 - r^2$ is pinned down, and that is exactly what the area difference needs. Whenever a shaded region is the difference of two circles or semicircles, look for a right triangle whose legs are a half-chord and the gap between the two boundaries: the Pythagorean theorem hands over $R^2 - r^2$ directly. The same trick solves the classic "annulus determined by a tangent chord" problem, where a chord of length $2a$ tangent to the inner circle gives ring area $\pi a^2$.

Since the answer cannot depend on $R$, choose a convenient case: let the small semicircle be tangent to $\overline{CD}$ at its midpoint and take $r = h$ so that $R^2 = 64 + r^2$. Setting $r = 6$ gives $R = 10$, and the area is $\frac{1}{2}\pi(100 - 36) = 32\pi$; setting $r = 8$ gives $R = 8\sqrt{2}$ and $\frac{1}{2}\pi(128 - 64) = 32\pi$. Two agreeing special cases confirm the general computation.

• Subtracting whole-circle areas instead of semicircle areas, which doubles the answer to 64pi, choice (E).
• Assuming the small semicircle is concentric with the large one; the figure shows it off-centre, and tangency alone already forces $r = h$ wherever it sits.
• Trying to solve for $R$ and $r$ separately; they are not determined, and only $R^2 - r^2$ is needed.