Competition · AMC preparation · step 4 of 4

AMC 8 · 2010 · #18

Grade 7 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
area-rectanglesarea-circlesratio-proportion identify-subproblemsconvert-to-algebra ↑ Prerequisites: area-rectanglesarea-circlesratio-proportion
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A decorative window is a rectangle ABCD with a semicircle attached to each of the two shorter sides (AB and CD). The side ratio is AD:AB = 3:2, and AB = 30 inches. Find the ratio of the rectangle's area to the combined area of the two semicircles.

Pick an answer.

(A)
2:3
(B)
3:2
(C)
$6:\pi$
(D)
$9:\pi$
(E)
$30:\pi$

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce Variables

The specific 30-inch length will cancel in a ratio, so Tool #5 (Introduce Variables) — let AB = 2k and AD = 3k — keeps the algebra clean and makes the cancellation visible. Tool #7 (Identify Subproblems) splits the figure into two pieces with separate area formulas: the rectangle (length × width) and the two semicircles, which combine into one full circle of radius k. Computing each area, then taking the ratio, makes the π stay where it belongs and the k² drop out.

1STEP 1

Name the side lengths

From AD:AB = 3:2, name the sides AB = 2k, AD = 3k and keep k symbolic — the 30 will cancel.

AB = 2k, AD = 3k
2STEP 2

Find the rectangle's area

Rectangle area = length × width = 3k × 2k = 6k².

Area_rect = AD × AB = 3k × 2k = 6k²
3STEP 3

Find the semicircle radius

Each semicircle's diameter is the short side AB = 2k, so its radius is r = k.

r = AB/2 = 2k/2 = k
4STEP 4

Combine the two semicircles

Two equal semicircles glue into one circle of radius k, so their combined area is π k².

Area_semis = π r² = π k²
5STEP 5

Take the ratio of areas

Divide 6k² by π k²; the k² cancels to give the ratio 6:π — choice (C).

Area_rect/Area_semis = 6k²/(π k²) = 6/π → 6:π → (C)
Answer
6:π
Plug the actual numbers back in: AB = 30 → k = 15, so Area_rect = 6 · 15² = 1350 in² and Area_semis = π · 15² = 225π in². The ratio 1350225π\frac{1350}{225π} = 6π\frac{6}{π} matches. Numerically, 6π\frac{6}{π} ≈ 1.91, so the rectangle is a bit less than twice the combined semicircles — that fits the picture: the rectangle is tall (45 × 30) and the semicircles together make a circle of radius 15, whose area is clearly smaller than the rectangle.
💡Key takeaway

This AMC 8 problem only needs the Grade 7 circle-area formula A = π r² plus a Grade 6 "use a letter for the side length" trick — the 30 inches was a red herring!

  • Name the side lengths
  • Find the rectangle's area
  • Find the semicircle radius
  • Combine the two semicircles
  • Take the ratio of areas

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