1982 AMC 12 Problem 4

Attempt Problem 4 of the 1982 AMC 12 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 1982 AMC 12 solutions, or check the answer key.

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

4.

The perimeter of a semicircular region, measured in centimeters, is numerically equal to its area, measured in square centimeters. The radius of the semicircle, measured in centimeters, is
π\pi
2π\frac2\pi
11
12\frac12
4π+2\frac4\pi+2
Answer: E
Concepts:circumferenceperimeterareaalgebraic manipulation
Small hint:
Include the diameter in the perimeter of the region
Big hint:
Set πr+2r\pi r+2r equal to πr22\frac{\pi r^2}{2}
Solution:
For r>0,r\gt0, equality of perimeter and area gives πr+2r=πr22.\pi r+2r=\frac{\pi r^2}{2}. Dividing by rr and solving yields r=2(π+2)π=2+4π.r=\frac{2(\pi+2)}{\pi}=2+\frac{4}{\pi}.
Therefore, the correct answer is E.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.