AMC 8 · 2008 · #25
Grade 7 geometry-2d
Pick an answer.
AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The black area is not one shape — it is the innermost disk plus two rings. Tool #7 (Identify Subproblems) splits the work into computing each black piece separately, then adding them. Tool #15 (Visualize) helps read the picture: starting from the outside, the colors go white, black, white, black, white, black. That tells us exactly which radii bound each black region. Once the black total and the design total are in hand, the percent is a single division.
Square each of the six radii 2, 4, 6, 8, 10, 12 with A = π r² to get the disk areas.
Grade 7 area-of-a-circle formula applied six times. Squaring even numbers is easy: r² goes 4, 16, 36, 64, 100, 144.
7.G.B.4Identify SubproblemsReading outward the colors go white, black, white, black, white, black, so the black is the inner disk plus two rings.
Tool #15: trace the rings from the outside and label them W, B, W, B, W, B. Each ring's area is the outer disk minus the inner disk.
7.G.B.4Organize Information In More WaysAdd the three black pieces: (100π - 64π) + (36π - 16π) + 4π = 60π.
Subtract, then add — Grade 7 arithmetic with the common factor π kept as is.
7.NS.A.3Identify SubproblemsThe whole design is the outer circle 144π, so the black share is = .
π cancels top and bottom. Divide 60 and 144 by their common factor 12 to get .
7.NS.A.3Identify SubproblemsAs a percent, ≈ 41.7%, whose closest choice is 42.
Grade 6 percent: multiply the fraction by 100%. 41.7% is closest to 42 on the list.
6.RP.A.3Identify SubproblemsDon't try to measure the black shape all at once. Cut it into a small disk and two rings, find each piece with π r², and the percent falls out.