AMC 10 · 2003 · #16
Grade 8 geometry-2d
Pick an answer.
The shaded piece has a jagged boundary made of four arcs, so measuring it head-on is awkward. Tool #16 (Count the Complement) flips the job: the shaded area is just the whole large semicircle minus the white region the small semicircles cover, and both of those are easy round shapes. To get the white area we still have to handle the overlaps, so tool #7 (Identify Subproblems) breaks the covered region into semicircles minus their shared lenses, and each lens into a triangle plus two circular slivers. Tool #1 (Draw a Diagram) with coordinates on AB keeps track of which semicircles actually overlap.
Set coordinates and the two basic areas
With the origin at the middle, the big area is 2π and each small one is π/2.
A semicircle is just half a circle, so its area is half of π r².
7.G.B.4Draw A DiagramRewrite the shaded area as a subtraction
The shaded part is the big one minus the white region the small ones cover.
It is easier to find the shaded strip by removing the round white part from the whole than to chase its wavy edge directly.
7.G.B.6Change Focus Count The ComplementSee which small semicircles overlap
The two outer ones only touch, but each meets the middle in a lens.
When shapes overlap, adding their areas counts the shared part twice, so subtract each overlap once.
Adding the semicircle areas counts each overlap twice, so every overlap must be subtracted once.
▸ Why?
Adding two regions counts whatever lies in both of them twice, so taking the overlap out once restores an honest total.
▸ Why?
The white cover is its pieces put together, so once each is counted exactly once the total is the area it really covers.
Measure one overlap lens
An equilateral triangle plus two sixty-degree slivers makes one lens π/3 - √3/4.
Two radius-1 circles whose centers are 1 apart make an equilateral triangle, which pins every angle to 60°.
8.G.B.7Identify SubproblemsAdd up the white region
Subtracting both lenses from the tripled area gives white as 5π/6 + √3/2.
Once each overlap is removed exactly once, the leftover is the true area the semicircles cover.
7.G.B.6Identify SubproblemsSubtract to get the shaded area
Subtracting that from the big semicircle leaves 7π/6 - √3/2, choice (E).
The shaded area is whatever is left of the big semicircle after the white cover is taken away.
7.G.B.6Change Focus Count The ComplementWhen a region has a messy edge, measure the whole minus the easy leftover, and remember overlapping shapes share area you must subtract once.
- Set coordinates and the two basic areas
- Rewrite the shaded area as a subtraction
- See which small semicircles overlap
- Measure one overlap lens
- Add up the white region
- Subtract to get the shaded area
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