AMC 8 · 2013 · #20

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

area-circlespythagorean-theoremreflection-symmetry identify-subproblems ↑ Prerequisites: pythagorean-theoremarea-circles

📏 Medium solution 💡 3 insights

Problem

A 1 × 2 rectangle sits inside a semicircle so that its longer side (length 2) lies flat on the diameter. Find the area of the semicircle.

Pick an answer.

(A)

$\frac{\pi}{2}$

(B)

$\frac{2\pi}{3}$

(C)

$\pi$

(D)

$\frac{4\pi}{3}$

(E)

$\frac{5\pi}{3}$

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The problem is pure 2D geometry, so Tool #1 (Draw a Diagram) is the right opener: a quick sketch — semicircle, rectangle, center of diameter — exposes a right triangle whose hypotenuse is the radius. Tool #7 (Identify Subproblems) then splits the work into two clean steps: (1) find r² from that right triangle (Pythagorean theorem), and (2) plug r² into the semicircle area formula. We do not need algebra (Tool #13) because the rectangle's symmetry hands us the right triangle directly.

1STEP 1

Center the diameter at the origin; by symmetry the top corners land on the arc at (-1, 1) and (1, 1).

Bottom: (-1,0), (1,0); Top on arc: (-1,1), (1,1)

2STEP 2

The radius from the origin to (1, 1) is the hypotenuse of a right triangle with legs 1 and 1.

legs = 1 and 1; hypotenuse = r

3STEP 3

By the Pythagorean theorem, r² = 1² + 1² = 2 — and the area formula needs only r², not r itself.

r² = 1² + 1² = 2

4STEP 4

Plug r² = 2 into the semicircle area $\frac{1}{2}$ π r²: $\frac{1}{2}$ π (2) = π → (C).

Area =

\frac{1}{2}

π r² =

\frac{1}{2}

π (2) = π → (C)

Answer

Sanity check the size. The diameter is 2r = 2√(2) ≈ 2.83, comfortably bigger than the rectangle's base of 2 — good, the rectangle fits. The semicircle's area π ≈ 3.14 must be larger than the rectangle's area 1 × 2 = 2, and it is (3.14 ＞ 2). It must also be smaller than the bounding 2√(2) × √(2) ≈ 4 rectangle that would contain the semicircle, and it is (3.14 ＜ 4). The answer (C) π sits in the right window.

💡Key takeaway

Draw the picture, find the right triangle hiding inside, and the Pythagorean theorem plus the circle-area formula do the rest — a Grade 8 idea applied to a tidy Grade 5 setup.

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