Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1982 AHSME Problems/Problem 4"
(Added solution and navbox)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
−
==Problem==
+
==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
 
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
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
−
by mathhayden (bad)
+
Let <math>r</math> be the radius of the semicircle.  Then its perimeter is <math>2r + \pi r = r(2 + \pi)</math>, while its area is <math>\frac{1}{2}\pi r^2</math>.  So:
  
−
perimeter=2r+pi2r=2r(pi+1)
+
<math>r(2 + \pi) = \frac{1}{2}\pi r^2</math>
  
−
area=[pi(r^2)]/2
+
<math>2 + \pi = \frac{1}{2}\pi r</math>
  
−
make them equal & solve for r:
+
<math>r = \frac{2}{\pi}(2 + \pi) = \boxed{(\mathbf{E})\ \frac{4}{\pi} + 2}</math>
−
pi(r^2)/2=2r(pi+1)
+
 
−
pi(r^2)=4r(pi+1)
+
-j314andrews
−
(r^2)=4r[(pi+1)/pi]
+
 
−
r=4(pi+1)/pi
+
==See Also==
−
=<math>4\pi+4/pi</math>   Ans.
+
{{AHSME box|year=1982|num-b=3|num-a=5}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 20:54, 29 June 2025

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

See Also

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. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_4&oldid=252859"