AMC 8 · 2018 · #15

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference

area-circlesratio-proportionformula-substitution area-differenceidentify-subproblems ↑ Prerequisites: area-circlesfraction-arithmetic

📏 Short solution 💡 3 insights 📊 Diagram

Problem

In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{1}{3}$

(C)

$\frac{1}{2}$

(D)

(E)

$\frac{\pi}{2}$

AMC 8 2018 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A large circle contains two smaller circles inside it. Each small circle's diameter equals the large circle's radius (so the two small circles fit snugly along a diameter of the large one). The two small circles together have area $1$ square unit. Find the area of the shaded part (the large circle minus the two small ones).

Givens: Two small congruent circles sit inside one large circle; Diameter of a small circle $=$ radius of the large circle; Combined area of the two small circles $= 1$ square unit; Shaded region $=$ (large circle) $-$ (two small circles); Answer choices: (A) $\frac{1}{4}$, (B) $\frac{1}{3}$, (C) $\frac{1}{2}$, (D) $1$, (E) $\frac{\pi}{2}$

Unknowns: The area, in square units, of the shaded region

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem

The shaded region is a compound shape, so Tool #7 (Subproblems) is the natural lead: shaded $=$ (large circle area) $-$ (two small circles' area). To make the radius relationship visible, Tool #1 (Diagram) — label the small radius $r$ and notice the large radius is $R = 2r$. Tool #9 (Easier Related Problem) is the safety net: instead of carrying $\pi$ symbolically, observe that the small circles' total area is *given* as $1$, so we only need to compare the large area to that given number — which turns the problem into the simple question "how many times bigger is the large circle than the two small ones combined?"

Execute — Answer: D

#1 Draw a Diagram 4.MD.A.1 Step 1

Draw and label.
Let $r$ be the radius of each small circle.
The picture says "a diameter of a small circle equals the radius of the large circle." A small diameter is $2r$, so the large radius is $R = 2r$.

$$R = 2r$$

💡 Labeling parts of a figure and using diameter $=2 \times$ radius is a Grade 4 measurement-units idea.

#7 Identify Subproblems 7.G.B.4 Step 2

Write each region's area using the circle-area formula $A = \pi r^2$.
The two small circles together have area $2 \pi r^2$.
The large circle has area $\pi R^2 = \pi (2r)^2 = 4 \pi r^2$.

$$A_{\text{small,total}} = 2\pi r^2,\quad A_{\text{large}} = \pi(2r)^2 = 4\pi r^2$$

💡 Knowing the circle-area formula $\pi r^2$ is the Grade 7 standard for circles.

#7 Identify Subproblems 6.EE.A.3 Step 3

Use the subproblem split: shaded $=$ (large circle) $-$ (two small circles).
Substituting the expressions in $r$, the shaded area is $4\pi r^2 - 2\pi r^2 = 2\pi r^2$ — exactly the *same* expression as the combined area of the two small circles.

$$A_{\text{shaded}} = 4\pi r^2 - 2\pi r^2 = 2\pi r^2$$

💡 Combining like terms ($4\pi r^2 - 2\pi r^2 = 2\pi r^2$) is the Grade 6 "equivalent expressions" move.

#9 Solve an Easier Related Problem 6.EE.B.5 Step 4

Match the expression to the given number.
The problem tells us the two small circles' combined area equals $1$, i.e.
$2 \pi r^2 = 1$.
From Step 3, the shaded area is *also* $2 \pi r^2$ — so it must equal $1$ too.

$$A_{\text{shaded}} = 2\pi r^2 = 1 \;\Rightarrow\; \textbf{(D)}\ 1$$

💡 Recognizing that two expressions with the same letters must have the same value is the Grade 6 idea of "a value that makes the equation true."

[1] #1 4.MD.A.1 Draw and label. Let $r$ be the radius of each small circle. The picture says "a

[2] #7 7.G.B.4 Write each region's area using the circle-area formula $A = \pi r^2$. The two sm

[3] #7 6.EE.A.3 Use the subproblem split: shaded $=$ (large circle) $-$ (two small circles). Sub

[4] #9 6.EE.B.5 Match the expression to the given number. The problem tells us the two small cir

Review

Reasonableness: Sanity check with pictures: the large circle has radius $2r$, so its area is $4$ times the area of one small circle (radius doubles $\Rightarrow$ area quadruples). One small circle has area $\tfrac{1}{2}$ (since two of them total $1$), so the large circle has area $4 \times \tfrac{1}{2} = 2$. Shaded $= 2 - 1 = 1$. ✓ Choice (D). Choice (E) $\tfrac{\pi}{2}$ is the classic trap for forgetting that the given $1$ already absorbed $\pi$ into a number.

Alternative: Tool #9 (Easier Related Problem) all the way: pick a friendly value, say $r = 1$. Then one small circle has area $\pi$, two small circles have area $2\pi$, and the large circle has area $\pi \cdot 2^2 = 4\pi$. Shaded $= 4\pi - 2\pi = 2\pi$, which is exactly the same as the two-small-circles total. So if the two small circles are *re-scaled* to combined area $1$, the shaded region scales the same way and is also $1$. Same answer, no $\pi$ left in the result.

CCSS standards used (min grade 7)

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units (Using the diameter–radius relation (diameter $= 2 \times$ radius) to translate the picture's statement into $R = 2r$.)
7.G.B.4 Know the formulas for area and circumference of a circle (Writing each circle's area with $A = \pi r^2$ — combined small circles $= 2\pi r^2$ and large circle $= \pi (2r)^2 = 4\pi r^2$.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Simplifying the shaded-area expression $4\pi r^2 - 2\pi r^2$ to the equivalent expression $2\pi r^2$.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Recognizing that the shaded area expression $2\pi r^2$ and the given combined-small-area $2\pi r^2 = 1$ are the same quantity, so the shaded area equals $1$.)

⭐ This AMC 8 problem only needs Grade 7 circle-area formula $A = \pi r^2$ — and the cool trick that the shaded ring and the two small circles end up with the exact same expression, so the answer is just the $1$ you were already given!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

More like this