Difference between revisions of "1982 AHSME Problems/Problem 4"
Mathhayden (talk | contribs) (Created page with "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 =<math>...") |
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| + | ==Problem== | ||
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| + | 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 | ||
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| + | <math>\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</math> | ||
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| + | ==Solution== | ||
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by mathhayden (bad) | by mathhayden (bad) | ||
Revision as of 13:57, 23 July 2022
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
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.
Ans.
