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.

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5.

A circle of diameter 11 is removed from a 2×32 \times 3 rectangle. Which whole number is closest to the area of the shaded region that remains?

11

22

33

44

55

Answer: E
Concepts:circle areaarea decompositionestimation

Difficulty rating: 820

Solution:

The rectangle has area 6,6, and the removed circle has area π(12)2=π40.79.\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4} \approx 0.79.

So the shaded region has area 60.795.2,6 - 0.79 \approx 5.2, whose closest whole number is 5.5.

Thus, the correct answer is E .

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