1992 AJHSME Problems/Problem 24

Problem

Four circles of radius $3$ are arranged as shown. Their centers are the vertices of a square. The area of the shaded region is closest to

$[asy] fill((3,3)--(3,-3)--(-3,-3)--(-3,3)--cycle,lightgray); fill(arc((3,3),(0,3),(3,0),CCW)--(3,3)--cycle,white); fill(arc((3,-3),(3,0),(0,-3),CCW)--(3,-3)--cycle,white); fill(arc((-3,-3),(0,-3),(-3,0),CCW)--(-3,-3)--cycle,white); fill(arc((-3,3),(-3,0),(0,3),CCW)--(-3,3)--cycle,white); draw(circle((3,3),3)); draw(circle((3,-3),3)); draw(circle((-3,-3),3)); draw(circle((-3,3),3)); draw((3,3)--(3,-3)--(-3,-3)--(-3,3)--cycle); [/asy]$

$\text{(A)}\ 7.7 \qquad \text{(B)}\ 12.1 \qquad \text{(C)}\ 17.2 \qquad \text{(D)}\ 18 \qquad \text{(E)}\ 27$

Solution

The side of the square of made up of two radii, and the area of the square is $(3+3)^2=36$. The unshaded region of the square is composed of four quarter circles, and has the area of one circle, which is $9\pi$. The area of the shaded region is $36-9\pi$ which less than $36-9(3)=9$, closest to $\boxed{\text{(A)}\ 7.7}$.