Difference between revisions of "1953 AHSME Problems/Problem 12"
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== Solution == | == Solution == | ||
| − | 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{( | + | 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{(B) }\frac{4}{9}}</math>. |
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==See Also== | ==See Also== | ||
Latest revision as of 14:04, 25 June 2024
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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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
