AMC 8 · 2010 · #18

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-rectanglesarea-circlesratio-proportion identify-subproblemsconvert-to-algebra ↑ Prerequisites: area-rectanglesarea-circlesratio-proportion

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A decorative window is made up of a rectangle with semicircles at either end. The ratio of $AD$ to $AB$ is $3:2$ . And $AB$ is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircles?

Pick an answer.

(A)

2:3

(B)

3:2

(C)

$6:\pi$

(D)

$9:\pi$

(E)

$30:\pi$

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Toolkit + CCSS Solution

Understand

Restated: A decorative window is a rectangle $ABCD$ with a semicircle attached to each of the two shorter sides ($AB$ and $CD$). The side ratio is $AD:AB = 3:2$, and $AB = 30$ inches. Find the ratio of the rectangle's area to the combined area of the two semicircles.

Givens: $ABCD$ is a rectangle with $AD:AB = 3:2$; $AB = 30$ inches (so $AD = 45$ inches); Two semicircles, one on side $AB$ and one on side $CD$, each with diameter equal to that side; Answer choices: (A) $2:3$, (B) $3:2$, (C) $6:\pi$, (D) $9:\pi$, (E) $30:\pi$

Unknowns: The ratio $\dfrac{\text{area of rectangle } ABCD}{\text{combined area of the two semicircles}}$

Understand

Plan

Primary tool: #5 Introduce Variables

Secondary: #7 Identify Subproblems

The specific $30$-inch length will cancel in a ratio, so Tool #5 (Introduce Variables) — let $AB = 2k$ and $AD = 3k$ — keeps the algebra clean and makes the cancellation visible. Tool #7 (Identify Subproblems) splits the figure into two pieces with separate area formulas: the rectangle (length $\times$ width) and the two semicircles, which combine into one full circle of radius $k$. Computing each area, then taking the ratio, makes the $\pi$ stay where it belongs and the $k^2$ drop out.

Execute — Answer: C

#5 Introduce Variables 6.EE.A.2 Step 1

Introduce a variable for the side lengths.
The ratio $AD:AB = 3:2$ means we can write $AB = 2k$ and $AD = 3k$ for some positive $k$.
The $30$-inch value of $AB$ will turn out to be irrelevant for the ratio, so we keep $k$ symbolic.

$$AB = 2k, \quad AD = 3k$$

💡 Naming the unknown lengths with a single letter $k$ is the Grade 6 idea of writing an expression where a letter stands for a number.

#7 Identify Subproblems 6.G.A.1 Step 2

Compute the area of the rectangle using length $\times$ width.

$$\text{Area}_{\text{rect}} = AD \times AB = 3k \times 2k = 6k^{2}$$

💡 Multiplying the two side lengths of a rectangle is the Grade 6 area-of-a-polygon move.

#7 Identify Subproblems 7.G.B.4 Step 3

Find the radius of each semicircle.
Each semicircle sits on a short side of length $AB = 2k$, which is its diameter, so the radius is half of that.

$$r = \dfrac{AB}{2} = \dfrac{2k}{2} = k$$

💡 Diameter equals the side length, so the radius is half of it — Grade 7 circle basics.

#7 Identify Subproblems 7.G.B.4 Step 4

Combine the two semicircles.
Two semicircles with the same radius glue into one full circle of radius $k$, so their combined area is the area of that single circle.

$$\text{Area}_{\text{semis}} = \pi r^{2} = \pi k^{2}$$

💡 Two halves make a whole — the Grade 7 circle area formula $A = \pi r^{2}$ applies directly.

#5 Introduce Variables 6.RP.A.1 Step 5

Take the ratio of the two areas.
The $k^{2}$ factor cancels, confirming that the specific value $AB = 30$ inches never mattered.

$$\dfrac{\text{Area}_{\text{rect}}}{\text{Area}_{\text{semis}}} = \dfrac{6k^{2}}{\pi k^{2}} = \dfrac{6}{\pi} \;\Rightarrow\; 6:\pi \;\Rightarrow\; \textbf{(C)}$$

💡 Expressing one area as a multiple of another is the Grade 6 ratio-language idea: "for every $\pi$ of circle, there are $6$ of rectangle."

[1] #5 6.EE.A.2 Introduce a variable for the side lengths. The ratio $AD:AB = 3:2$ means we can

[2] #7 6.G.A.1 Compute the area of the rectangle using length $\times$ width.

[3] #7 7.G.B.4 Find the radius of each semicircle. Each semicircle sits on a short side of leng

[4] #7 7.G.B.4 Combine the two semicircles. Two semicircles with the same radius glue into one

[5] #5 6.RP.A.1 Take the ratio of the two areas. The $k^{2}$ factor cancels, confirming that the

Review

Reasonableness: Plug the actual numbers back in: $AB = 30 \Rightarrow k = 15$, so $\text{Area}_{\text{rect}} = 6 \cdot 15^{2} = 1350$ in$^{2}$ and $\text{Area}_{\text{semis}} = \pi \cdot 15^{2} = 225\pi$ in$^{2}$. The ratio $\dfrac{1350}{225\pi} = \dfrac{6}{\pi}$ matches. Numerically, $6/\pi \approx 1.91$, so the rectangle is a bit less than twice the combined semicircles — that fits the picture: the rectangle is tall ($45 \times 30$) and the semicircles together make a circle of radius $15$, whose area is clearly smaller than the rectangle.

Alternative: Tool #3 (Eliminate Possibilities) on the choices. Any answer that does not involve $\pi$ (choices A and B) cannot be right, because the semicircle area must contribute a $\pi$. That leaves (C) $6:\pi$, (D) $9:\pi$, (E) $30:\pi$. Choice (E) is the trap of using $AB = 30$ directly without dividing for the radius, and (D) is the trap of squaring $3$ from the $3:2$ side ratio. The honest computation above gives the coefficient $6 = 3 \times 2$ from $AD \cdot AB / r^{2}$, matching (C).

CCSS standards used (min grade 7)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Setting $AB = 2k$ and $AD = 3k$ so the ratio condition $3:2$ is built into the side-length expressions.)
6.G.A.1 Find the area of polygons by composing or decomposing into rectangles and triangles (Computing the rectangle's area as $AD \times AB = 6k^{2}$.)
6.RP.A.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship (Expressing the final answer as the ratio $6 : \pi$ between the two areas.)
7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems (Combining the two semicircles into one full circle of radius $k$ and applying $A = \pi r^{2}$ to get $\pi k^{2}$.)

⭐ This AMC 8 problem only needs the Grade 7 circle-area formula $A = \pi r^{2}$ plus a Grade 6 "use a letter for the side length" trick — the $30$ inches was a red herring!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

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