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

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

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

Area_rect = AD × AB = 3k × 2k = 6k²

3STEP 3

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

r =

\frac{AB}{2}

\frac{2k}{2}

= k

4STEP 4

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

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

\frac{Area\_rect}{Area\_semis}

\frac{6k²}{π k²}

\frac{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

\frac{1350}{225π}

\frac{6}{π}

matches. Numerically,

\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!

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