AMC 8 · 2011 · #25
Grade 8 geometry-2dProblem
A circle with radius 1 is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?
Pick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A circle of radius $1$ sits inside a larger square (touching all four sides) and outside a smaller square (whose four corners are on the circle). The picture has two shaded pieces: the part of the circle outside the small square, and the part of the large square outside the circle. We need the ratio (circle-shaded area) : (area between the two squares), and pick the answer choice closest to that ratio.
Givens: Circle radius $r = 1$; Large square is circumscribed about the circle (each side touches the circle); Small square is inscribed in the circle (each vertex lies on the circle); Answer choices: (A) $\tfrac{1}{2}$, (B) $1$, (C) $\tfrac{3}{2}$, (D) $2$, (E) $\tfrac{5}{2}$
Unknowns: The fraction closest to $\dfrac{\text{circle's shaded area}}{\text{area between the two squares}}$
Understand
Restated: A circle of radius $1$ sits inside a larger square (touching all four sides) and outside a smaller square (whose four corners are on the circle). The picture has two shaded pieces: the part of the circle outside the small square, and the part of the large square outside the circle. We need the ratio (circle-shaded area) : (area between the two squares), and pick the answer choice closest to that ratio.
Givens: Circle radius $r = 1$; Large square is circumscribed about the circle (each side touches the circle); Small square is inscribed in the circle (each vertex lies on the circle); Answer choices: (A) $\tfrac{1}{2}$, (B) $1$, (C) $\tfrac{3}{2}$, (D) $2$, (E) $\tfrac{5}{2}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities
The picture has three nested shapes — large square, circle, small square — so Tool #7 (Identify Subproblems) splits the work into three clean area calculations and a final ratio. Tool #1 (Draw a Diagram) is built into the problem's figure: reading off that the circle's diameter equals the large square's side and the small square's diagonal is what unlocks every dimension. With a clean numerical ratio in hand, Tool #3 (Eliminate Possibilities) finishes the job by comparing one decimal to the five answer choices.
Execute — Answer: A
7.G.B.4 Step 1 - Read the two key length facts from the figure.
- The circle is inscribed in the large square, so the large square's side equals the circle's diameter: $\text{side} = 2r = 2$.
- The small square is inscribed in the circle, so its diagonal equals the diameter: $\text{diagonal} = 2r = 2$.
💡 Tool #1: the figure is doing work for us — touching points and corners are how the circle's radius transfers to the squares.
3.MD.C.7 Step 2 - Subproblem 1: area of the large square.
- A square's area is side squared.
💡 Grade 3 "area = side $\times$ side" — the easiest of the three pieces.
8.G.B.7 Step 3 - Subproblem 2: area of the small square.
- Let its side be $s$.
- The diagonal of a square splits it into two right triangles with legs $s, s$ and hypotenuse $s\sqrt{2}$.
- So $s\sqrt{2} = 2$, meaning $s^2 = 2$ — which is exactly the area.
💡 Pythagorean theorem on the half-square turns the diagonal into the side-squared, which is the area itself.
7.G.B.4 Step 4 Subproblem 3: area of the circle.
💡 Grade 7 circle-area formula. The $r = 1$ trick makes the answer just $\pi$.
3.MD.C.7 Step 5 - Assemble the two shaded regions.
- The circle's shaded part is the circle minus the small square (the corners of the small square are inside the circle).
- The region between the squares is the large square minus the small square.
💡 Subtracting nested areas is the standard "region between shapes" move from Grade 3 area reasoning.
6.RP.A.3 Step 6 Form the ratio and estimate using $\pi \approx 3.14$.
💡 A ratio of two areas is a Grade 6 rate/ratio computation — same skill as miles per hour.
6.RP.A.3 Step 7 - Compare $0.57$ to the answer choices and pick the closest.
- The gaps are $|0.57 - 0.5| = 0.07$ and $|0.57 - 1| = 0.43$; every other choice is even farther away.
💡 Tool #3: "closest to" problems become a distance check between one number and five candidates.
7.G.B.4 Read the two key length facts from the figure. The circle is inscribed in the la 3.MD.C.7 Subproblem 1: area of the large square. A square's area is side squared. 8.G.B.7 Subproblem 2: area of the small square. Let its side be $s$. The diagonal of a s 7.G.B.4 Subproblem 3: area of the circle. 3.MD.C.7 Assemble the two shaded regions. The circle's shaded part is the circle minus th 6.RP.A.3 Form the ratio and estimate using $\pi \approx 3.14$. 6.RP.A.3 Compare $0.57$ to the answer choices and pick the closest. The gaps are $|0.57 - Review
Reasonableness: Sanity check the size: the circle ($\pi \approx 3.14$) is a bit bigger than the small square ($2$) but smaller than the large square ($4$), which matches the picture. The circle's shaded crescents add up to a thin sliver — about $1.14$ — and the four corner pieces between the squares add up to $2$. So the shaded circle area is about half the between-squares area, and $\tfrac{1}{2}$ matches.
Alternative: Tool #16 (Change Focus): rewrite the ratio as $\dfrac{\pi - 2}{2} = \dfrac{\pi}{2} - 1$. Since $\pi/2 \approx 1.57$, the ratio is about $0.57$ — same conclusion, but no decimal subtraction needed. This also makes it obvious that the answer must be a little above $\tfrac{1}{2}$ and far below $1$, so (A) wins without even checking the other choices.
CCSS standards used (min grade 8)
3.MD.C.7Relate area to multiplication; find area of rectangles by tiling and multiplying side lengths (Computing the large square's area as $2 \times 2 = 4$ and forming the "between squares" area as $4 - 2 = 2$ by subtracting nested regions.)7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Reading the circle's diameter from the figure ($d = 2r = 2$) and computing the circle's area as $\pi r^2 = \pi$.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Turning the small square's diagonal ($2$) into its area: $s\sqrt{2} = 2$ gives $s^2 = 2$, which is the area.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Forming the ratio $\dfrac{\pi - 2}{2} \approx 0.57$ and selecting the closest of the five fractional choices.)
⭐ Three nested shapes break into three Grade 3-8 area facts, and a $\pi \approx 3.14$ estimate turns the ratio into a single decimal you can match to the closest answer choice.
⭐ Three nested shapes break into three Grade 3-8 area facts, and a $\pi \approx 3.14$ estimate turns the ratio into a single decimal you can match to the closest answer choice.
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