2025 AMC 10A Problem 10

P1-10~4 minPrint

Circles Add construction lines/points (drop altitudes, extend segments, connect centers)Cut the figure into known shapes (triangles, rectangles, sectors)

A semicircle has diameter $\overline{AB}$ and chord $\overline{CD}$ of length $16$ parallel to $\overline{AB}$ . A smaller semicircle with diameter on $\overline{AB}$ and tangent to $\overline{CD}$ is cut from the larger semicircle, as shown below.

What is the area of the resulting figure, shown shaded?

Show solution without answering

Key insight

The small radius equals the height of the chord above AB. A right triangle with legs 8 and that height gives big radius squared minus small radius squared = 64.

Solution

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
$r = h.$

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 $h$ .

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

Why this works

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.

Alternative approach

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 $\tfrac12 \pi (128 - 64) = 32\pi$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)
Introduction to Geometry, Chapter 12: Circles and Angles