Art of Problem Solving

1970 AMC 12 Problems/Problem 2: Difference between revisions

CHONGWON (talk | contribs)
CHONGWON (talk | contribs)
Line 3: Line 3:
A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is
A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is


<math> \mathrm{(A) \ } pi/4\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } h\qquad \mathrm{(D) \ } 2h</math>
<math>\mathrm{(A)}\ \frac 2pi\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78</math>
 
<math>\mathrm{(E) \ }  h^3</math>


== Solution ==
== Solution ==

Revision as of 11:34, 9 January 2008

Problem

A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is

$\mathrm{(A)}\ \frac 2pi\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78$

Solution