2025 AMC 10A Problem 10

Below is the professionally curated solution for Problem 10 of the 2025 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10A solutions, or check the answer key.

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Concepts:circle area Pythagorean Theorem

Difficulty rating: 1440

10.

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

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

$16\pi$

$24\pi$

$32\pi$

$48\pi$

$64\pi$

Solution:

Let $O$ be the center on $AB$ and $P$ the midpoint of chord $CD.$ Set $r = OP$ for the small radius and $R = OD$ for the large one. Since $PD = 8,$ the Pythagorean theorem in triangle $OPD$ gives $R^2 - r^2 = 64.$ The shaded area is the big semicircle minus the small one: $\tfrac12\pi R^2 - \tfrac12\pi r^2 = \tfrac12\pi(R^2 - r^2) = 32\pi.$ Therefore, the answer is C.

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