AMC 8 · 2010 · #16

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition

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

📏 Medium solution 💡 3 insights

Problem

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

Pick an answer.

(A)

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

(B)

$\sqrt{\pi}$

(C)

$\pi$

(D)

$2\pi$

(E)

$\pi^{2}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Area of the square equals area of the circle; Square area formula: $s^2$, where $s$ is the side length; Circle area formula: $\pi r^2$, where $r$ is the radius; Answer choices: (A) $\tfrac{\sqrt{\pi}}{2}$, (B) $\sqrt{\pi}$, (C) $\pi$, (D) $2\pi$, (E) $\pi^2$

Unknowns: The ratio $\dfrac{s}{r}$ (square side $:$ circle radius)

Understand

Restated: 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?

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #1 Draw a Diagram, #6 Guess and Check

The ratio $s/r$ 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 $\pi$?" 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 $s/r$, and only one squares back to $\pi$.

Execute — Answer: B

#1 Draw a Diagram 3.MD.C.7 Step 1

Draw the picture.
Sketch a square of side $s$ next to a circle of radius $r$.
Label the square's area as $s^2$ and the circle's area as $\pi r^2$.
Setting these equal is the whole problem in one picture.

$$s^2 = \pi r^2$$

💡 The Grade 3 area formula for a square (side $\times$ side) and the standard circle area formula give two expressions that the problem forces to match.

#9 Solve an Easier Related Problem 7.G.B.4 Step 2

Shrink to an easier case.
Since the ratio $s/r$ is the same for every valid pair, pick the simplest circle: $r = 1$.
Then the circle's area is $\pi \cdot 1^2 = \pi$, so the square must also have area $\pi$.

$$r = 1 \;\Rightarrow\; \pi r^2 = \pi \;\Rightarrow\; s^2 = \pi$$

💡 The Grade 7 circle-area formula $\pi r^2$ collapses to $\pi$ when $r = 1$, removing the variable $r$ from the problem.

#9 Solve an Easier Related Problem 8.EE.A.2 Step 3

Solve for the side length.
The square's area is the side squared, so the side is the (positive) square root of the area.

$$s^2 = \pi \;\Rightarrow\; s = \sqrt{\pi}$$

💡 Taking a positive square root to undo a square is the Grade 8 "use square root symbols to represent solutions" move.

#6 Guess and Check 6.RP.A.1 Step 4

Form the ratio.
With $r = 1$ and $s = \sqrt{\pi}$, the ratio of side length to radius is just $s$ itself.

$$\dfrac{s}{r} = \dfrac{\sqrt{\pi}}{1} = \sqrt{\pi} \;\Rightarrow\; \textbf{(B)}$$

💡 Writing a ratio of two measured lengths is the Grade 6 ratio definition — and the answer matches choice (B).

[1] #1 3.MD.C.7 Draw the picture. Sketch a square of side $s$ next to a circle of radius $r$. La

[2] #9 7.G.B.4 Shrink to an easier case. Since the ratio $s/r$ is the same for every valid pair

[3] #9 8.EE.A.2 Solve for the side length. The square's area is the side squared, so the side is

[4] #6 6.RP.A.1 Form the ratio. With $r = 1$ and $s = \sqrt{\pi}$, the ratio of side length to r

Review

Reasonableness: Sanity-check the size. $\sqrt{\pi} \approx \sqrt{3.14} \approx 1.77$, 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 $4r^2$), 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: $\pi \approx 3.14$ and $2\pi \approx 6.28$ would make the square far bigger than the bounding $2r$ square, which is impossible.

Alternative: Tool #6 (Guess and Check) on the choices: square each candidate and see which gives $\pi$. (A) $(\tfrac{\sqrt{\pi}}{2})^2 = \tfrac{\pi}{4}$, (B) $(\sqrt{\pi})^2 = \pi$ ✓, (C) $\pi^2$, (D) $4\pi^2$, (E) $\pi^4$. Only (B) squares back to $\pi$, matching $s^2/r^2 = \pi$ from $s^2 = \pi r^2$.

CCSS standards used (min grade 8)

3.MD.C.7 Relate area to multiplication; find area of rectangles with whole-number side lengths (Writing the square's area as side $\times$ side $= s^2$ — the Grade 3 area-of-a-rectangle idea applied to a square.)
7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems (Writing the circle's area as $\pi r^2$ and evaluating it at $r = 1$ to get area $\pi$ for the easier case.)
8.EE.A.2 Use square root and cube root symbols to represent solutions to equations of the form $x^2 = p$ (Solving $s^2 = \pi$ by taking the positive square root to get $s = \sqrt{\pi}$.)
6.RP.A.1 Understand the concept of a ratio and use ratio language to describe a relationship (Reading the question as "ratio of side to radius" and forming $s : r = \sqrt{\pi} : 1$ to identify the answer.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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