AMC 10 · 2009 · #6

Grade 7 geometry-2d

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

area-circlesarea-differencefraction-arithmetic identify-subproblemscomplementary-counting ↑ Prerequisites: area-circles

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

A circle of radius $2$ is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded?

Pick an answer.

(A)

$\ \frac{1}{2}$

(B)

$\ \frac{\pi}{6}$

(C)

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

(D)

$\ \frac{2}{3}$

(E)

$\ \frac{3}{\pi}$

AMC 10 2009 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.

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Toolkit + CCSS Solution

Understand

Restated: A circle of radius $2$ sits snugly inside a semicircle, touching the flat straight edge and the curved arc. The region inside the semicircle but outside the circle is shaded. Find what fraction of the semicircle's area is shaded.

Givens: The inscribed circle has radius $2$; The circle is inscribed in the semicircle: it rests on the diameter and touches the arc; The shaded region is the part inside the semicircle but outside the circle; Answer choices: (A) $\tfrac{1}{2}$, (B) $\tfrac{\pi}{6}$, (C) $\tfrac{2}{\pi}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{\pi}$

Unknowns: The ratio (shaded area) / (semicircle area)

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #16 Change Focus / Count the Complement

The one fact the picture hides is the semicircle's radius. Reading the figure carefully (Tool #1) unlocks it: the circle stands on the flat edge and its top just reaches the arc, so its full diameter fits between them and equals the semicircle's radius. Once both radii are known, the area question splits into two clean pieces (Tool #7): the semicircle's area and the circle's area. The shaded part is not computed directly but as what is left over after removing the circle (Tool #16), which turns the fraction into a simple subtraction and cancellation.

Execute — Answer: A

#1 Draw a Diagram 7.G.B.4 Step 1

Read the semicircle's radius from the figure

The circle rests on the straight diameter and its highest point touches the arc.
Its center is $2$ above the diameter (one radius up), and its top is another $2$ above the center, so the top of the circle is $4$ above the diameter.
That distance from the flat edge straight up to the arc is exactly the semicircle's radius.
So the semicircle has radius $4$, meaning the circle's whole diameter equals the semicircle's radius.

$$R_{\text{semi}} = 2 + 2 = 4$$

💡 The circle stacks bottom-to-top from the flat edge to the arc, so its diameter is the big radius.

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

Area of the semicircle

A semicircle is half a full circle.
With radius $4$, the full circle would have area $\pi \cdot 4^2 = 16\pi$, so the semicircle is half of that.

$$A_{\text{semi}} = \tfrac{1}{2}\pi (4)^2 = \tfrac{1}{2}\cdot 16\pi = 8\pi$$

💡 Use the circle-area formula on radius $4$, then take half because only a semicircle is drawn.

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

Area of the inscribed circle

The inscribed circle has radius $2$, so apply the same area formula directly.

$$A_{\text{circ}} = \pi (2)^2 = 4\pi$$

💡 A radius-$2$ circle has area $4\pi$ straight from $\pi r^2$.

#16 Change Focus / Count the Complement 6.RP.A.3 Step 4

Subtract, then form the fraction

The shaded region is everything inside the semicircle except the circle, so its area is the semicircle's area minus the circle's area: $8\pi - 4\pi = 4\pi$.
The fraction shaded is this over the whole semicircle.
The $\pi$ cancels and the fraction reduces cleanly.

$$\dfrac{A_{\text{shaded}}}{A_{\text{semi}}} = \dfrac{8\pi - 4\pi}{8\pi} = \dfrac{4\pi}{8\pi} = \dfrac{1}{2}\ \textbf{(A)}$$

💡 The circle eats exactly half the semicircle's area, so what is left is the other half.

[1] #1 7.G.B.4 The circle rests on the straight diameter and its highest point touches the arc.

[2] #7 7.G.B.4 A semicircle is half a full circle. With radius $4$, the full circle would have

[3] #7 7.G.B.4 The inscribed circle has radius $2$, so apply the same area formula directly.

[4] #16 6.RP.A.3 The shaded region is everything inside the semicircle except the circle, so its

Review

Reasonableness: The circle's area $4\pi$ is exactly half the semicircle's area $8\pi$, so the shaded leftover must also be $4\pi$ — the two pieces split the semicircle evenly. A fraction of $\tfrac{1}{2}$ is one of the listed choices and matches the picture, where the white circle visibly fills about half the region. Note every $\pi$ cancels, so the answer is a clean rational number, which fits choice (A) rather than the $\pi$-containing options (B), (C), (E).

Alternative: Skip the actual areas and compare with a scale factor. The semicircle's radius ($4$) is twice the circle's radius ($2$). Doubling a radius multiplies a full circle's area by $2^2 = 4$, so a full circle of radius $4$ has $4$ times the circle's area. But only a semicircle is drawn — half of that full circle — giving $\tfrac{4}{2} = 2$ times the small circle's area. So the semicircle equals two of the inscribed circles: one fills the white part, the other is the shaded part, again $\tfrac{1}{2}$.

CCSS standards used (min grade 7)

7.G.B.4 Know the formulas for area and circumference of a circle (Relating the inscribed circle's diameter to the semicircle's radius, and computing the areas $\tfrac{1}{2}\pi(4)^2 = 8\pi$ and $\pi(2)^2 = 4\pi$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Forming the part-to-whole ratio $\tfrac{4\pi}{8\pi}$ and reducing it to $\tfrac{1}{2}$.)

⭐ The circle's diameter equals the semicircle's radius, so the circle covers exactly half the semicircle and the shaded leftover is the other half, $\tfrac{1}{2}$.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Read the semicircle's radius from the figure

Area of the semicircle

Area of the inscribed circle

Subtract, then form the fraction

Review

CCSS standards used (min grade 7)

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