Difference between revisions of "1982 AHSME Problems/Problem 4"

(Solution)
 
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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>
 
<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>
  
==Solution==
+
Option E
 
 
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>4\pi+4/pi</math>  Ans.
 

Latest revision as of 02:16, 31 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

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

Option E