Competition · AMC preparation · step 4 of 4

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 12\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

Find figure A's shaded area

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

Find figure B's shaded area

Figure B: four small circles each π · (12\frac{1}{2})² = π4\frac{π}{4} total 4 · π4\frac{π}{4} = π, so B also leaves 4 - π.

Area_B = 2² - 4 · π (1/2)² = 4 - 4 · π/4 = 4 - π
3STEP 3

Find figure C's shaded area

Figure C: the inscribed square (diagonal 2) has area 222\frac{2²}{2} = 2, so circle π · 1² = π minus it leaves π - 2.

Area_C = π · 1² - 2²/2 = π - 2
4STEP 4

Compare the three areas

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.

  • Find figure A's shaded area
  • Find figure B's shaded area
  • Find figure C's shaded area
  • Compare the three areas

A parent dashboard for the family lives at sensimlab.com.

More like this

Same problem type. Tags show what each one shares with this problem.