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AMC 8 · 2003 · #22

Grade 7 geometry-2d
Archetype Compound Area by Decomposition / Difference
area-rectanglesarea-circlespythagorean-theoremspatial-visualization area-differenceidentify-subproblemssystematic-enumeration ↑ Prerequisites: area-rectanglesarea-circlespythagorean-theorem
📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The following figures are composed of squares and circles. Which figure has a shaded region with largest area?

Pick an answer.

(A)
A only
(B)
B only
(C)
C only
(D)
both A and B
(E)
all are equal
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Toolkit + CCSS Solution

Understand

Restated: Three figures sit side by side, each built from a $2$ cm square and circles. Figure A is a $2 \times 2$ square with one inscribed circle (radius $1$) removed. Figure B is a $2 \times 2$ square with four small circles (radius $\tfrac{1}{2}$) 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?

Givens: Figure A: $2 \times 2$ square with one circle of radius $1$ cut out; Figure B: $2 \times 2$ square with four circles of radius $\tfrac{1}{2}$ cut out (one in each quadrant); Figure C: circle of radius $1$ with a square whose diagonal equals the diameter $2$ cut out; All three side lengths/diameters are $2$ cm; Answer choices: (A) A only, (B) B only, (C) C only, (D) both A and B, (E) all are equal

Unknowns: Which figure (or figures) has the largest shaded area

Understand

Restated: Three figures sit side by side, each built from a $2$ cm square and circles. Figure A is a $2 \times 2$ square with one inscribed circle (radius $1$) removed. Figure B is a $2 \times 2$ square with four small circles (radius $\tfrac{1}{2}$) 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?

Givens: Figure A: $2 \times 2$ square with one circle of radius $1$ cut out; Figure B: $2 \times 2$ square with four circles of radius $\tfrac{1}{2}$ cut out (one in each quadrant); Figure C: circle of radius $1$ with a square whose diagonal equals the diameter $2$ cut out; All three side lengths/diameters are $2$ cm; Answer choices: (A) A only, (B) B only, (C) C only, (D) both A and B, (E) all are equal

Plan

Primary tool: #1 Draw a Diagram

Secondary: #9 Solve an Easier Problem

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 $a - b\pi$ or $b\pi - a$ means a single end-of-problem comparison decides the winner; no decimals are needed until the last step.

Execute — Answer: C

#9 Solve an Easier Problem 7.G.B.4 Step 1
  • Figure A: the $2 \times 2$ square has area $2^2 = 4$.
  • The inscribed circle has radius $1$, so its area is $\pi \cdot 1^2 = \pi$.
  • The shaded region is the square minus the circle.
$$\text{Area}_A = 2^2 - \pi \cdot 1^2 = 4 - \pi$$

💡 Grade 7 area-of-a-circle formula: $\pi r^2$ with $r = 1$ gives $\pi$. Then a single subtraction finishes A.

#9 Solve an Easier Problem 7.G.B.4 Step 2
  • Figure B: same $2 \times 2$ square (area $4$), but now four small circles of radius $\tfrac{1}{2}$ are cut out.
  • Each small circle has area $\pi \cdot \left(\tfrac{1}{2}\right)^2 = \tfrac{\pi}{4}$.
  • Four of them remove $4 \cdot \tfrac{\pi}{4} = \pi$ in total — the same amount of white as in A.
$$\text{Area}_B = 2^2 - 4 \cdot \pi \left(\tfrac{1}{2}\right)^2 = 4 - 4 \cdot \tfrac{\pi}{4} = 4 - \pi$$

💡 Halving the radius cuts the area to one quarter, and there are four circles — the factor of $4$ in count cancels the factor of $\tfrac{1}{4}$ in area. A and B tie.

#1 Draw a Diagram 6.G.A.1 Step 3
  • Figure C: a circle of radius $1$ with a square removed.
  • The diameter of the circle equals the diagonal of the square, both $2$.
  • For a square whose diagonal is $d$, the area is $\tfrac{d^2}{2}$, so the inscribed square has area $\tfrac{2^2}{2} = 2$.
  • The shaded region is the circle minus the square.
$$\text{Area}_C = \pi \cdot 1^2 - \tfrac{2^2}{2} = \pi - 2$$

💡 Draw the diagonal of the inscribed square — it splits the square into two right triangles with legs along the diagonals, and the diagonal-formula $\tfrac{d_1 d_2}{2}$ gives area $2$ in one step.

#1 Draw a Diagram 7.NS.A.3 Step 4
  • Compare the three shaded areas.
  • A and B are both $4 - \pi$; C is $\pi - 2$.
  • Using $\pi \approx 3.14$, the first is $4 - 3.14 = 0.86$ and the second is $3.14 - 2 = 1.14$.
  • So C is largest.
$$4 - \pi \approx 0.86 \;\;<\;\; \pi - 2 \approx 1.14 \;\Rightarrow\; \textbf{(C)}$$

💡 Only the final comparison needs a decimal for $\pi$. The diagram-side picture (a slim ring of leftover circle in C versus thin corners in A and B) matches the inequality.

[1] #9 7.G.B.4 Figure A: the $2 \times 2$ square has area $2^2 = 4$. The inscribed circle has r
[2] #9 7.G.B.4 Figure B: same $2 \times 2$ square (area $4$), but now four small circles of rad
[3] #1 6.G.A.1 Figure C: a circle of radius $1$ with a square removed. The diameter of the circ
[4] #1 7.NS.A.3 Compare the three shaded areas. A and B are both $4 - \pi$; C is $\pi - 2$. Usin

Review

Reasonableness: Sanity-check with the picture. In A and B the white shapes use up the same total area ($\pi$ 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 $4 - \pi$. In C the inscribed square only covers half of the bounding $2 \times 2$ square (area $2$ vs $4$), so the circle's interior $\pi \approx 3.14$ minus $2$ leaves about $1.14$ — bigger than the $\approx 0.86$ left over in A or B. The answer (C) is the only one consistent with all three exact areas.

Alternative: Tool #6 (Guess and Check) with a numerical estimate: use $\pi \approx 3.14$ from the start. Then $\text{Area}_A = \text{Area}_B \approx 0.86$ and $\text{Area}_C \approx 1.14$. The same ranking pops out, and the tie between A and B is visible immediately — leading to choice (C).

CCSS standards used (min grade 7)

  • 7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems (Computing the area of each circle ($\pi$ for the radius-$1$ circles in A and C, $\tfrac{\pi}{4}$ for each radius-$\tfrac{1}{2}$ circle in B) so each figure's shaded area can be written as a simple expression in $\pi$.)
  • 6.G.A.1 Find the area of polygons by composing into rectangles or decomposing into triangles (Finding the area of the inscribed square in C from its diagonal: split it into two right triangles whose legs lie along the diagonals to get area $\tfrac{d^2}{2} = 2$.)
  • 7.NS.A.3 Solve real-world and mathematical problems involving the four operations with rational numbers (Comparing $4 - \pi$ with $\pi - 2$ at the end using the approximation $\pi \approx 3.14$ to decide which shaded area is largest.)

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

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

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