2025 AMC 10A Problem 10

Call the big radius $R$ and the small radius $r$. The shaded area is the big half circle minus the small half circle:
$$\tfrac12 \pi R^2 - \tfrac12 \pi r^2 = \tfrac12 \pi \left(R^2 - r^2\right).$$
So we do not need $R$ and $r$ separately. We only need $R^2 - r^2$.

1. Find the small radius. Let $h$ be the height of chord $\overline{CD}$ above $\overline{AB}$. The small semicircle sits on $\overline{AB}$, so its top point is $r$ above $\overline{AB}$. "Tangent to $\overline{CD}$" means that top point just touches $\overline{CD}$. So
$$r = h.$$

1. Draw a right triangle in the big semicircle. Let $O$ be the center of the big semicircle, the midpoint of $\overline{AB}$. Draw the radius $OC$, which has length $R$. Also draw the line from $O$ straight up to $\overline{CD}$, at a right angle to it; its length is $h$.

1. Read off the sides. A line from the center that meets a chord at a right angle cuts the chord in half. So it hits $\overline{CD}$ at its middle, $8$ from $C$ (half of $16$). The right triangle has legs $8$ and $h$, and hypotenuse $R$.

4. Use the Pythagorean theorem.
$$R^2 = 8^2 + h^2 = 64 + r^2, \qquad\text{so}\qquad R^2 - r^2 = 64.$$

5. Finish the area.
$$\tfrac12 \pi \cdot 64 = 32\pi.$$

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

The problem does not tell you either radius, and the small semicircle could slide left or right. That is a hint that only a combination like $R^2 - r^2$ matters. When a shaded region is one circle (or half circle) minus another, look for a right triangle with a half-chord as one leg: the Pythagorean theorem gives $R^2 - r^2$ in one step.

Since the answer cannot depend on the sizes, pick numbers that fit. Suppose the chord is $6$ above $\overline{AB}$, so $r = 6$. The right triangle with legs $8$ and $6$ has hypotenuse $10$, so $R = 10$. The shaded area is
$$\tfrac12 \pi (10^2 - 6^2) = \tfrac12 \pi \cdot 64 = 32\pi.$$
Trying a different height, say $r = 8$, gives $R^2 = 64 + 64 = 128$ and the same $\tfrac12 \pi (128 - 64) = 32\pi$.

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