Difference between revisions of "1970 AHSME Problems/Problem 2"

Problem

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

$\text{(A) } \frac{4}{\pi}\quad \text{(B) } \frac{\pi}{\sqrt{2}}\quad \text{(C) } \frac{4}{1}\quad \text{(D) } \frac{\sqrt{2}}{\pi}\quad \text{(E) } \frac{\pi}{4}$

Solution

Let's say the circle has a circumference ( or perimeter of $4\pi$). Since the perimeter of the square is the same as the perimeter of the circle, the side length of the square is $\pi$. That means that the area of the square is $\pi^2$. The area of the circle is $4\pi$. So the area of the circle over the area of the square is $\frac{4\pi}{\pi^2} \Rightarrow \frac{4}{\pi} \Rightarrow$ $\fbox{A}$

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