1982 AHSME Problems/Problem 4

Problem

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

$\textbf{(A)} \ \pi \qquad \textbf{(B)} \ \frac{2}{\pi} \qquad \textbf{(C)} \ 1 \qquad \textbf{(D)} \ \frac{1}{2}\qquad \textbf{(E)} \ \frac{4}{\pi}+2$

Solution

by mathhayden (bad)

perimeter=2r+pi2r=2r(pi+1)

area=[pi(r^2)]/2

make them equal & solve for r: pi(r^2)/2=2r(pi+1) pi(r^2)=4r(pi+1) (r^2)=4r[(pi+1)/pi] r=4(pi+1)/pi

=   Ans.

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1982 AHSME Problems/Problem 4

Problem

Solution