Problem Solving

This problem is from the Quant Review
Figure shows 3 identical circles that are tangent to each other. If the area of the shaded region is 64Root3-32PI, what is the radius of each circle?
OG 8


This is how I attacked the problem.
First I noticed that the triangle formed was an equilateral triangle. I know if I find the side of that I can multiply by 1/2 and get the radius of the circle.
I used the formula for the equilateral and take a look at the equation given. The equilateral equation is (S^2ROOT3)/4. The 64 root 3 part looks familiar so I multiply the 64 by 4 and get 256. I know that 16^2 is 256 so r must be 1/2*16 which is 8.

Is this the best way to do this problem?

The figure shown consists of three identical circles that are tangent to each other. If the area of the shaded region (center space where the three circles do not touch) is (64)(square root 3) - 32 pi, what is the radius of each circle?

a. 4
b. 8
c. 16
d. 24
e. 32

OA: B

Shaded region = triangle - 3 circle sectors = 64√3 - 32�

Looking at the equation above, we can see that the 3 circle sectors = 32�.
Since the triangle is equilateral, each of its angles is 60 degrees. Since 60/360 = 1/6, each circle sector is 1/6 the area of each circle.
Thus, the 3 sectors = 3*1/6 = 1/2 each circle area.
Since the 3 sectors = 32�, circle area = 64�.
Thus, �r² = 64�, and r=8.

The correct answer is B.

let the radius of a circle be x
since the triangle is equilateral, its angle is equal to 60degree
the bottom of the triangle is 2x, the diagonal is x*sqroot3
the area of the triangle =2x*x*sqroot3/2=(x^2)*sqroot3
the area of the sector of a cirle =(60/360)*pi*x^2=(1/6)*pi*x^2

so we have -
(x^2)*sqroot3 -3* (1/6)*pi*x^2=64sqroot3-32pi
x^2(sqroot3-(1/2)*pi*x^2)=32(2sqroot3-pi)
x^2=32*2
x=8