1982 AHSME Problems/Problem 4

Problem 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

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

Solution

Let $r$ be the radius of the semicircle. Then its perimeter is $2r + \pi r = r(2 + \pi)$ , while its area is $\frac{1}{2}\pi r^2$ . So:

$r(2 + \pi) = \frac{1}{2}\pi r^2$

$2 + \pi = \frac{1}{2}\pi r$

$r = \frac{2}{\pi}(2 + \pi) = \boxed{(\mathbf{E})\ \frac{4}{\pi} + 2}$

-j314andrews

1982 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

Page

Toolbox

Search

1982 AHSME Problems/Problem 4

Problem 4

Solution

See Also