Difference between revisions of "1997 AJHSME Problems/Problem 7"

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If the circle has radius 4, it has diameter 8.  Thus, the smallest square that contains it has side length 8, and area <math>8\times8=64</math>.
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==Problem==
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The area of the smallest square that will contain a circle of radius 4 is
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<math>\text{(A)}\ 8 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128</math>
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==Solution==
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Draw a square circumscribed around the circle.  (Alternately, the circle is inscribed in the square.)  If the circle has radius <math>4</math>, it has diameter <math>8</math>.  Two of the diameters of the circle will run parallel to the sides of the square.  Thus, the smallest square that contains it has side length <math>8</math>, and area <math>8\times8=64</math>.
 
<math>\boxed{\textbf{(D)}}</math>
 
<math>\boxed{\textbf{(D)}}</math>
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== See also ==
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{{AJHSME box|year=1997|num-b=6|num-a=8}}
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* [[AJHSME]]
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* [[AJHSME Problems and Solutions]]
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* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 23:26, 4 July 2013

Problem

The area of the smallest square that will contain a circle of radius 4 is

$\text{(A)}\ 8 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$

Solution

Draw a square circumscribed around the circle. (Alternately, the circle is inscribed in the square.) If the circle has radius $4$, it has diameter $8$. Two of the diameters of the circle will run parallel to the sides of the square. Thus, the smallest square that contains it has side length $8$, and area $8\times8=64$. $\boxed{\textbf{(D)}}$

See also

1997 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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
All AJHSME/AMC 8 Problems and Solutions

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