AMC 8 · 2025 · #18

Grade 7 geometry-2d

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

area-circlessimilar-figuresarea-trianglesperfect-squares easier-related-problemidentify-subproblemsarea-difference ↑ Prerequisites: area-circlesarea-triangleslinear-equations-one-var

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The circle shown below on the left has a radius of 1 unit. The region between the circle and the inscribed square is shaded. In the circle shown on the right, one quarter of the region between the circle and the inscribed square is shaded. The shaded regions in the two circles have the same area. What is the radius $R$ , in units, of the circle on the right?

Pick an answer.

(A)

$\sqrt{2}$

(B)

(C)

$2\sqrt{2}$

(D)

(E)

$4\sqrt{2}$

AMC 8 2025 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: Two circles each contain an inscribed square (corners on the circle). The left circle has radius $1$, and the whole region inside the circle but outside the square is shaded. The right circle has radius $R$, and only one quarter of that same in-between region is shaded. The two shaded areas are equal — find $R$.

Givens: Left circle: radius $= 1$, with an inscribed square; Left shaded region = (entire circle) $-$ (inscribed square); Right circle: radius $= R$, with an inscribed square; Right shaded region = $\tfrac{1}{4}$ of [(entire circle) $-$ (inscribed square)]; The two shaded areas are equal; Answer choices: (A) $\sqrt{2}$, (B) $2$, (C) $2\sqrt{2}$, (D) $4$, (E) $4\sqrt{2}$

Unknowns: The radius $R$ of the right circle

Understand

Plan

Primary tool: #7 Identify Subproblems

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

The shaded "in-between" region is awkward as a single shape, but tool #7 (Subproblems) turns it into a clean difference: (circle area) $-$ (inscribed-square area). Tool #1 (Diagram) — specifically drawing the two diagonals of the inscribed square — splits the square into four right isoceles triangles whose legs are the radius, so we get the square's area as $2r^2$ without needing Pythagoras. Tool #9 (Easier Problem) is the key insight: both circles have the same shape, so both "between" regions follow the same formula $(\pi - 2)r^2$. That reduces the whole problem to solving $(\pi - 2)(1)^2 = \tfrac{1}{4}(\pi - 2)R^2$, which is just $R^2 = 4$.

Execute — Answer: B

#1 Draw a Diagram 6.G.A.1 Step 1

Find the area of the inscribed square when the circle has radius $r$.
Draw the two diagonals of the square; they cross at the center of the circle and split the square into $4$ congruent right isoceles triangles, each with two legs of length $r$ (a radius).

$$\text{square area} = 4 \times \tfrac{1}{2} \cdot r \cdot r = 2r^{2}$$

💡 Cutting the square into $4$ right triangles whose legs are radii is the Grade 6 "decompose a polygon into triangles to find its area" idea — no Pythagoras needed.

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

Write the area of the region between the circle and its inscribed square (the "annulus-like" shape).
It is (circle area) $-$ (square area).
Using $r$ for the radius:

$$\text{between area}(r) = \pi r^{2} - 2r^{2} = (\pi - 2)\, r^{2}$$

💡 The between region isn't a standard shape, but subtracting the square from the circle (Grade 7 area formula $\pi r^{2}$) makes it a single clean expression.

#9 Solve an Easier Related Problem 7.G.B.4 Step 3

Apply that formula to each picture.
The left picture shades the WHOLE between region (radius $1$); the right picture shades ONE QUARTER of the between region (radius $R$).

$$\text{left shaded} = (\pi - 2)(1)^{2} = \pi - 2 \qquad \text{right shaded} = \tfrac{1}{4}(\pi - 2)\, R^{2}$$

💡 Because both circles have the same shape (just resized), the between-region formula $(\pi - 2)r^{2}$ works for both — that is the Grade 7 "area scales with $r^{2}$" pattern.

#7 Identify Subproblems 6.EE.B.7 Step 4

The problem says the two shaded areas are equal.
Set them equal and solve for $R$.
The factor $(\pi - 2)$ is positive ($\pi \approx 3.14 > 2$), so we can divide both sides by it.

$$\pi - 2 = \tfrac{1}{4}(\pi - 2)\, R^{2} \;\Rightarrow\; 1 = \tfrac{R^{2}}{4} \;\Rightarrow\; R^{2} = 4 \;\Rightarrow\; R = 2$$

💡 Once we set up $1 = R^{2}/4$, finding $R$ is the Grade 6 "solve a one-step equation" skill ($R$ must be positive, so we take the positive root).

#7 Identify Subproblems 6.EE.B.7 Step 5

Match $R = 2$ to the answer choices.
That is choice (B).

$$R = 2 \;\Rightarrow\; \textbf{(B)}$$

💡 Reading the final value off the choice list is the closing step of any multiple-choice equation problem.

[1] #1 6.G.A.1 Find the area of the inscribed square when the circle has radius $r$. Draw the t

[2] #7 7.G.B.4 Write the area of the region between the circle and its inscribed square (the "a

[3] #9 7.G.B.4 Apply that formula to each picture. The left picture shades the WHOLE between re

[4] #7 6.EE.B.7 The problem says the two shaded areas are equal. Set them equal and solve for $R

[5] #7 6.EE.B.7 Match $R = 2$ to the answer choices. That is choice (B).

Review

Reasonableness: Intuition check: the right between-region only contributes one quarter of itself to the shaded area, so it has to be $4$ times bigger overall than the left between-region. Area scales with the square of the radius, so the radius only needs to be $\sqrt{4} = 2$ times bigger — not $4$ times. That rules out (D) $4$ and (E) $4\sqrt{2}$ immediately and lands exactly on (B) $2$, matching our algebra. A numerical sanity check: left shaded $= \pi - 2 \approx 1.14$; right shaded $= \tfrac{1}{4}(\pi - 2)(2)^{2} = \pi - 2 \approx 1.14$. They match.

Alternative: Tool #9 (Easier Related Problem) shortcut without any algebra: notice that the left and right pictures are similar (same shape, just resized by factor $R$). The whole between region on the right is therefore $R^{2}$ times as big as on the left. The right picture shades $\tfrac{1}{4}$ of that, so its shaded area is $\tfrac{R^{2}}{4}$ times the left shaded area. Setting that ratio equal to $1$ gives $R^{2} = 4$, so $R = 2$ — no need to compute $\pi - 2$ at all.

CCSS standards used (min grade 7)

6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing the inscribed square into $4$ right isoceles triangles with legs $r$ to get its area $2r^{2}$ without using the Pythagorean theorem.)
7.G.B.4 Know the formulas for area and circumference of a circle (Writing the circle area as $\pi r^{2}$ for both circles, so the between-region area can be expressed as $(\pi - 2)r^{2}$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving the equation $\pi - 2 = \tfrac{1}{4}(\pi - 2)R^{2}$ for the positive value $R = 2$.)

⭐ This AMC 8 problem only needs Grade 7 "area of a circle is $\pi r^{2}$" plus the idea that when you scale a shape, its area grows by the square of the scale — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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