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 squared is equal to $\pi$ which is irrational. Therefore, the answer is $\boxed{B}.$

~YJC64002776

1952 AHSC (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

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1952 AHSME Problems/Problem 1

Contents

Problem

Solution

Solution 2

See also