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

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==Problem==
 
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. The sum of the areas of the circles is:
 
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. The sum of the areas of the circles is:
  
 
<math>\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}</math>
 
<math>\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}</math>
  
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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>
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==Solution==
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Note the areas of these circles 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}}

Latest revision as of 00:38, 4 February 2020

Problem

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:

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

Solution

Note the areas of these circles 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 (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions


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