1953 AHSME Problems/Problem 12

Problem

The diameters of two circles are $8$ inches and $12$ inches respectively. The ratio of the area of the smaller to the area of the larger circle is:

$\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{4}{9} \qquad \textbf{(C)}\ \frac{9}{4} \qquad \textbf{(D)}\ \frac{1}{2}\qquad \textbf{(E)}\ \text{none of these}$

Solution

The area of a circle can be calculated as $\pi{r^2}$ where $r$ is the radius. We know that the radii of the circles are $4$ and $6$ inches (half the diameter) so the ratio of the area of the smaller to the area of the larger circle is $\frac{16\pi}{36\pi}=\boxed{\textbf{(C) }\frac{4}{9}}$.


See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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