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

 
Line 14: Line 14:
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 +
{{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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png