# Difference between revisions of "1952 AHSME Problems/Problem 1"

## Problem

If the radius of a circle is a rational number, its area is given by a number which is:

$\textbf{(A)\ } \text{rational} \qquad \textbf{(B)\ } \text{irrational} \qquad \textbf{(C)\ } \text{integral} \qquad \textbf{(D)\ } \text{a perfect square }\qquad \textbf{(E)\ } \text{none of these}$

## Solution

Let the radius of the circle be the common fraction $\frac{a}{b}.$ Then the area of the circle is $\pi \cdot \frac{a^2}{b^2}.$ Because $\pi$ is irrational and $\frac{a^2}{b^2}$ is rational, their product must be irrational. The answer is $\boxed{B}.$

## Solution 2

The phrasing of the problem makes it clear that the rule would apply to all rational radii. So let r be equal to one. $\pi$ times {1^2} is equal to $\pi$ which is irrational. Therefore, the answer is $\boxed{B}.$

~YJC64002776