Three congruent circles each of radius $4$ cm touches each other. What is the area (in $cm^{2}$) of the portion included between them?
- $8\pi$
- $16\sqrt {3} - 8\pi$
- $16\sqrt {3} - 4\pi$
- $16\sqrt {3} - 2\pi$
The centers of the three mutually touching circles form an equilateral triangle with side length equal to twice the radius, which is 8 cm. The area of this equilateral triangle is (sqrt(3)/4) * 8^2 = 16 * sqrt(3) square centimeters. The area of the three 60-degree sectors inside the triangle is 3 * (60/360) * pi * 4^2 = 8 * pi square centimeters, so the enclosed area is 16 * sqrt(3) - 8 * pi.
Connecting the centers of the three touching circles forms an equilateral triangle with a side length of 8 centimeters, as each side consists of two radii of 4 centimeters. Using the equilateral triangle area formula, the square root of 3 divided by 4 times side squared, the area is the square root of 3 divided by 4 times 8 squared, which is 16 times the square root of 3. Since the included portion is the triangle minus the sectors of the three circles, and the sectors form a semicircle of radius 4, the area is 16 times the square root of 3 minus 8 times pi.