Difference between revisions of "1953 AHSME Problems/Problem 27"

(Created page with "The radius of the first circle is <math>1</math> inch, that of the second <math>\frac{1}{2}</math> inch, that of the third <math>\frac{1}{4}</math> inch and so on indefinitely...")
 
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Note the areas of these triangles is <math>1\pi</math>, <math>\frac{\pi}{4}</math>, <math>\frac{\pi}{16}, \dots</math>. The sum of these areas will thus be <math>\pi\left(1+\frac{1}{4}+\frac{1}{16}+\dots\right)</math>. We use the formula for an infinite geometric series to get the sum of the areas will be <math>\pi\left(\frac{1}{1-\frac{1}{4}}\right)</math>, or <math>\frac{4\pi}{3}</math>.  <math>\boxed{D}</math>
 
Note the areas of these triangles is <math>1\pi</math>, <math>\frac{\pi}{4}</math>, <math>\frac{\pi}{16}, \dots</math>. The sum of these areas will thus be <math>\pi\left(1+\frac{1}{4}+\frac{1}{16}+\dots\right)</math>. We use the formula for an infinite geometric series to get the sum of the areas will be <math>\pi\left(\frac{1}{1-\frac{1}{4}}\right)</math>, or <math>\frac{4\pi}{3}</math>.  <math>\boxed{D}</math>
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==See Also==
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{{AHSME 50p box|year=1953|num-b=26|num-a=28}}
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{{MAA Notice}}

Revision as of 00:37, 4 February 2020

The radius of the first circle is $1$ inch, that of the second $\frac{1}{2}$ inch, that of the third $\frac{1}{4}$ inch and so on indefinitely. The sum of the areas of the circles is:

$\textbf{(A)}\ \frac{3\pi}{4} \qquad \textbf{(B)}\ 1.3\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ \frac{4\pi}{3}\qquad \textbf{(E)}\ \text{none of these}$

Note the areas of these triangles is $1\pi$, $\frac{\pi}{4}$, $\frac{\pi}{16}, \dots$. The sum of these areas will thus be $\pi\left(1+\frac{1}{4}+\frac{1}{16}+\dots\right)$. We use the formula for an infinite geometric series to get the sum of the areas will be $\pi\left(\frac{1}{1-\frac{1}{4}}\right)$, or $\frac{4\pi}{3}$. $\boxed{D}$

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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All AHSME Problems and Solutions


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