2004 AMC 8 Problem 25

Attempt Problem 25 of the 2004 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

25.

Two 4×44 \times 4 squares intersect at right angles, bisecting their intersecting sides, as shown. The circle’s diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?
16−4π16-4\pi
16−2π16-2\pi
28−4π28-4\pi
28−2π28-2\pi
32−2π32-2\pi
Answer: D
Concepts:circle areainclusion-exclusion
Small hint:
Find the area covered by the union of the two squares.
Big hint:
The circle diameter is the diagonal of the 2×22\times2 overlap square.
Solution:
The two 4×44\times4 squares have total area 32,32, but their overlap is a 2×22\times2 square with area 4.4. Thus the area covered by the union of the two squares is 32−4=28.32-4=28.
The circle’s diameter is the diagonal of that 2×22\times2 overlap square, so the diameter is 222\sqrt2 and the radius is 2.\sqrt2.
The circle area is π(2)2=2π,\pi(\sqrt2)^2=2\pi, so the shaded area is 28−2π.28-2\pi.
Thus, D is the correct answer.
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