AMC 8 · 2003 · #22

Grade 7 geometry-2d

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

area-rectanglesarea-circlespythagorean-theoremspatial-visualization area-differenceidentify-subproblemssystematic-enumeration ↑ Prerequisites: area-rectanglesarea-circlespythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Three figures sit side by side, each built from a 2 cm square and circles. Figure A is a 2 × 2 square with one inscribed circle (radius 1) removed. Figure B is a 2 × 2 square with four small circles (radius

\frac{1}{2}

) removed. Figure C is a circle (radius 1) with an inscribed square removed (the square's diagonal equals the diameter 2). Which figure has the largest shaded area?

Pick an answer.

(A)

A only

(B)

B only

(C)

C only

(D)

both A and B

(E)

all are equal

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The three figures are already labeled pictures, so Tool #1 (Draw a Diagram) lets us read off the shaded region of each as "outer shape minus inner shapes." Tool #9 (Solve an Easier Problem) splits one hard question — "which is biggest?" — into three small sub-problems: find the shaded area of A, then B, then C, each a quick subtraction. Keeping each area in exact form a - bπ or bπ - a means a single end-of-problem comparison decides the winner; no decimals are needed until the last step.

1STEP 1

Figure A: square area 2² = 4 minus the inscribed circle π · 1² = π leaves 4 - π.

Area_A = 2² - π · 1² = 4 - π

2STEP 2

Figure B: four small circles each π · ( $\frac{1}{2}$ )² = $\frac{π}{4}$ total 4 · $\frac{π}{4}$ = π, so B also leaves 4 - π.

Area_B = 2² - 4 · π (

\frac{1}{2}

)² = 4 - 4 ·

\frac{π}{4}

= 4 - π

3STEP 3

Figure C: the inscribed square (diagonal 2) has area $\frac{2²}{2}$ = 2, so circle π · 1² = π minus it leaves π - 2.

Area_C = π · 1² -

\frac{2²}{2}

= π - 2

4STEP 4

With π ≈ 3.14: 4 - π ≈ 0.86 is smaller than π - 2 ≈ 1.14, so the largest shaded area is figure C.

4 - π ≈ 0.86 ＜ π - 2 ≈ 1.14 → (C)

Answer

C only

Sanity-check with the picture. In A and B the white shapes use up the same total area (π in both, because four quarter-area circles equal one full circle), so the leftover shaded areas must match — that rules out (A) and (B) as winners and confirms the tie 4 - π. In C the inscribed square only covers half of the bounding 2 × 2 square (area 2 vs 4), so the circle's interior π ≈ 3.14 minus 2 leaves about 1.14 — bigger than the ≈ 0.86 left over in A or B. The answer (C) is the only one consistent with all three exact areas.

💡Key takeaway

Don't compare pictures by eye — write each shaded area as "outer minus inner." A and B both leave 4 - π, and C leaves π - 2. Since π - 2 is bigger than 4 - π, figure C wins.

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