Competition · AMC preparation · step 4 of 4
AMC 8 · 2025 · #18
Grade 7 geometry-2d
Pick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 2r² 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 (π - 2)r². That reduces the whole problem to solving (π - 2)(1)² = 1/4(π - 2)R², which is just R² = 4.
Find the inscribed square's area
Draw the square's two diagonals: they split it into 4 right isosceles triangles with legs equal to radius r, so its area is 2r².
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.
In a circle of radius r, the inscribed square has area 2r².
▸ Why?
Drawing the square's two diagonals cuts it into four triangles that meet at the center and fill the square with no gaps or overlaps, so the square's area is the sum of those four triangles' areas.
▸ Why?
In each of those four triangles, the two sides running from the center out to the square's corners each have length r, because the corners lie on the circle.
▸ Why?
Each triangle is right-angled where its two length-r sides meet, so each has area 1/2 r · r = 1/2r², and the four together give 4·1/2r² = 2r².
▸ Why?
The two diagonals cross at the center and split the turn there into four equal angles; two neighbors lie along one straight diagonal, so each of the four angles is half of a straight angle, a right angle.
▸ Why?
In each right triangle the two perpendicular length-r sides act as base and height, so its area is 1/2 · r · r = 1/2r².
Write the between area formula
The between-region is (circle area) - (square area), which cleans up to (π - 2) r².
The between region isn't a standard shape, but subtracting the square from the circle (Grade 7 area formula π r²) makes it a single clean expression.
7.G.B.4Identify SubproblemsApply it to both pictures
Left shades the whole region, giving π - 2; right shades a quarter, giving (π - 2) R².
Because both circles have the same shape (just resized), the between-region formula (π - 2)r² works for both — that is the Grade 7 "area scales with r²" pattern.
7.G.B.4Solve An Easier Related ProblemSet the shaded areas equal
Set them equal: π - 2 = (π - 2) R². Cancel the positive (π - 2) to get R² = 4, so R = 2.
Once we set up 1 = R²/4, finding R is the Grade 6 "solve a one-step equation" skill (R must be positive, so we take the positive root).
6.EE.B.7Identify SubproblemsMatch the radius to the choices
R = 2 matches answer choice (B).
Reading the final value off the choice list is the closing step of any multiple-choice equation problem.
6.EE.B.7Identify SubproblemsThis AMC 8 problem only needs Grade 7 "area of a circle is π r²" plus the idea that when you scale a shape, its area grows by the square of the scale — that you already know!
- Find the inscribed square's area
- Write the between area formula
- Apply it to both pictures
- Set the shaded areas equal
- Match the radius to the choices
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