AMC 10 · 2025 · #10

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference
area-circlesarea-differencechord-perpendicular-from-centerpythagorean-theorem identify-subproblems ↑ Prerequisites: area-circlespythagorean-theorem
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A large semicircle has diameter AB. Inside it a chord CD of length 16 runs parallel to AB. A smaller semicircle, with its flat diameter lying on AB and its curved top just touching CD, is removed from the large one. Find the area of the shaded region that is left.

Pick an answer.

(A)
$16\pi$
(B)
$24\pi$
(C)
$32\pi$
(D)
$48\pi$
(E)
$64\pi$

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

No radius is given a number, so I cannot compute either semicircle's area on its own. Tool #4 (Introduce a Variable) is the move: call the two radii R and r, write the shaded area in terms of them, and watch what the problem actually forces you to know. It turns out you never need R or r alone — only the single combination R² - r² — and that is exactly what one right triangle hands you. Tool #7 (Identify Subproblems) splits the shaded region into big-semicircle-minus-small-semicircle, and Tool #1 (Draw a Diagram) supplies the radius-to-chord right triangle that pins down R² - r².

1STEP 1

Split the shaded area

The shaded piece is the large semicircle minus the small one, and each semicircle is half of π times radius squared.

Shaded = (large semicircle) - (small semicircle)
2STEP 2

Name the two radii

Let R and r be the two radii. Factoring out π/2 leaves the shaded area depending only on R² - r², never on R and r separately.

Shaded = 1/2π R² - 1/2π r² = π/2(R² - r²)
3STEP 3

Tangency fixes the height of CD

The small semicircle tops out r above AB, and a tangent line grazes exactly that top point, so CD sits at height r.

height of CD above AB = r
4STEP 4

Right triangle from center to C

Drop a perpendicular from the center to CD: it cuts the chord into halves of 8 and has length r, so R² - r² = 64.

8² + r² = R² ⟹ R² - r² = 64
5STEP 5

Substitute and finish

Substituting R² - r² = 64 into the Step 2 expression kills both unknown radii and leaves 32π.

Shaded = π/2(R² - r²) = π/2 · 64 = 32π → (C)
Answer
32π
The final area 32π never used a specific value of R or r, which matches the problem never telling us where the small semicircle sits — good sign. Test the extreme case where the small semicircle shrinks to nothing (r → 0): then CD slides down onto AB, the chord of length 16 becomes the diameter, so R = 8 and the plain semicircle area is 1/2π (8)² = 32π. Same number, so the subtraction is consistent. And 32π is choice (C), squarely in the middle of the options.
💡Key takeaway

When no radius is given, name them with letters and watch for a combination like R² - r² that a single right triangle can hand you — you often never need the radii by themselves.

  • Split the shaded area
  • Name the two radii
  • Tangency fixes the height of CD
  • Right triangle from center to C
  • Substitute and finish

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