Competition · AMC preparation · step 4 of 4
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.
List the six disk areas
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 SubproblemsFind the black regions
Reading 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 black areas
Add the three black pieces: (100π - 64π) + (36π - 16π) + 4π = 60π.
Subtract, then add — Grade 7 arithmetic with the common factor π kept as is.
Adding the three black pieces together gives a total black area of 60π square inches.
▸ Why?
The black part is three separate regions that never overlap — the small center disk and two rings — so the whole black area is exactly those three areas added together.
▸ Why?
Cutting a region into pieces that leave no gaps and do not overlap means the pieces add back to make the whole.
▸ Why?
Each black ring's area is the disk out to its outer edge minus the smaller disk it wraps around.
▸ Why?
A full disk splits with no overlap into the inner disk plus the ring around it, so the ring is what is left once the inner disk is taken away.
▸ Why?
The inner disk and the ring around it together fill the whole outer disk, with nothing missing and nothing counted twice.
▸ Why?
Removing the inner disk's area is the reverse of adding it, so subtracting it recovers the ring exactly.
▸ Why?
Each disk's area is fixed by its radius through the circle rule, giving 4π, 16π, 36π, 64π, 100π for the disks that bound the black pieces.
▸ Why?
A disk of radius r always covers an area of π times r squared, so its size is decided by the radius alone.
Divide by the whole area
The 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 5/12.
7.NS.A.3Identify SubproblemsConvert to a percent
As 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.
- List the six disk areas
- Find the black regions
- Add the black areas
- Divide by the whole area
- Convert to a percent
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