Difference between revisions of "1953 AHSME Problems/Problem 12"

(Created page with "== Problem == The diameters of two circles are <math>8</math> inches and <math>12</math> inches respectively. The ratio of the area of the smaller to the area of the larger c...")
 
 
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The area of a circle can be calculated as <math>\pi{r^2}</math> where <math>r</math> is the radius. We know that the radii of the circles are <math>4</math> and <math>6</math> inches (half the diameter) so the ratio of the area of the smaller to the area of the larger circle is <math>\frac{16\pi}{36\pi}=\boxed{\textbf{(C) }\frac{4}{9}}</math>.
 
The area of a circle can be calculated as <math>\pi{r^2}</math> where <math>r</math> is the radius. We know that the radii of the circles are <math>4</math> and <math>6</math> inches (half the diameter) so the ratio of the area of the smaller to the area of the larger circle is <math>\frac{16\pi}{36\pi}=\boxed{\textbf{(C) }\frac{4}{9}}</math>.
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==See Also==
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{{AHSME 50p box|year=1953|num-b=11|num-a=13}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:49, 1 April 2017

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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All AHSME Problems and Solutions

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