1992 AMC 8 Problem 5

Attempt Problem 5 of the 1992 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 1992 AMC 8 solutions, or check the answer key.

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

5.

A circle of diameter 11 is removed from a 2×32 \times 3 rectangle, as shown. Which whole number is closest to the area of the shaded region?
11
22
33
44
55
Answer: E
Concepts:circle areaarea decompositionestimation
Small hint:
The rectangle has area 2×3=62 \times 3 = 6; subtract the area of the circle
Big hint:
The circle has radius 12,\dfrac12, so its area is π(12)2=π4,\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4}, a little less than 11
Solution:
The rectangle has area 6,6, and the removed circle has area π(12)2=π4≈0.79.\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4} \approx 0.79.
So the shaded region has area 6−0.79≈5.2,6 - 0.79 \approx 5.2, whose closest whole number is 5.5.
Thus, the correct answer is E .
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