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]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:49, 1 April 2017
Problem
The diameters of two circles are inches and inches respectively. The ratio of the area of the smaller to the area of the larger circle is:
Solution
The area of a circle can be calculated as where is the radius. We know that the radii of the circles are and inches (half the diameter) so the ratio of the area of the smaller to the area of the larger circle is .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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