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>. | + | ==Problem== |
| + | The area of the smallest square that will contain a circle of radius 4 is | ||
| + | |||
| + | <math>\text{(A)}\ 8 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128</math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | 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> | ||
| + | |||
| + | == See also == | ||
| + | {{AJHSME box|year=1997|num-b=6|num-a=8}} | ||
| + | * [[AJHSME]] | ||
| + | * [[AJHSME Problems and Solutions]] | ||
| + | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:26, 5 July 2013
Problem
The area of the smallest square that will contain a circle of radius 4 is
Solution
Draw a square circumscribed around the circle. (Alternately, the circle is inscribed in the square.) If the circle has radius 
, it has diameter 
. 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 , and area 
.
See also
| 1997 AJHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
