1997 AJHSME Problems/Problem 7

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 $8\times8=64$ . $\boxed{\textbf{(D)}}$

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

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1997 AJHSME Problems/Problem 7

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