AMC 8 · 2010 · #16

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

area-rectanglesarea-circlesratio-proportionexponents identify-subproblemsconvert-to-algebra ↑ Prerequisites: area-rectanglesarea-circlesexponents

📏 Medium solution 💡 3 insights

Problem

A square and a circle have exactly the same area. What is the ratio of the square's side length to the circle's radius?

Pick an answer.

(A)

$\frac{\sqrt{\pi}}{2}$

(B)

$\sqrt{\pi}$

(C)

$\pi$

(D)

$2\pi$

(E)

$\pi^{2}$

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

The ratio $\frac{s}{r}$ does not depend on the actual size, so Tool #9 (Easier Related Problem) lets us nail r = 1 and turn the question into: "what side length gives a square with area π?" That single concrete case answers the ratio without abstract variables. Tool #1 (Draw a Diagram) makes the equal-area condition visible — sketch a unit circle and a square of the same area side by side. Tool #6 (Guess and Check) is the multiple-choice safety net: each choice predicts a numerical value for $\frac{s}{r}$ , and only one squares back to π.

1STEP 1

Draw a square of side s beside a circle of radius r, then set their two areas equal.

s² = π r²

2STEP 2

Pick the simplest circle, r = 1, so its area is π and the square must also have area π.

r = 1 → π r² = π → s² = π

3STEP 3

A square's area is its side squared, so the side is the positive square root of the area: s = √(π).

s² = π → s = √(π)

4STEP 4

With r = 1 and s = √(π), the ratio of side to radius is just s itself, √(π) — choice (B).

\frac{s}{r}

\frac{√(π)}{1}

= √(π) → (B)

Answer

√(π)

Sanity-check the size. √(π) ≈ √(3.14) ≈ 1.77, so the square's side is a bit less than twice the circle's radius. That fits the picture: a circle of radius r fits inside a square of side 2r (area 4r²), so an equal-area square must be a bit smaller than 2r on a side. 1.77r lands right in that window — not bigger than 2r, not absurdly small. The other choices fail this test: π ≈ 3.14 and 2π ≈ 6.28 would make the square far bigger than the bounding 2r square, which is impossible.

💡Key takeaway

When a ratio doesn't depend on size, pick the easiest case (r = 1) and the whole problem shrinks to one square root!

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