2012 AMC 10B Problem 16

1. First the circles. Each has area $\pi \cdot 2^2 = 4\pi$, so the three circles give $12\pi$.
2. The rest of the shaded region is the curved gap in the middle. To measure it, connect the three centers.
3. Two touching circles of radius $2$ have centers $2 + 2 = 4$ apart. So the centers form an equilateral triangle with side $4$. Its area is
$$\frac{\sqrt3}{4} \cdot 4^2 = 4\sqrt3.$$
4. This triangle is made of the middle gap plus three pie slices, one inside each circle. Each slice has the triangle's $60^\circ$ corner angle and radius $2$.
5. Three $60^\circ$ slices make $180^\circ$, which is half a circle. Half of a circle with area $4\pi$ is $2\pi$.
6. So the gap is the triangle minus the slices: $4\sqrt3 - 2\pi$.
7. Total shaded area: circles plus gap,
$$12\pi + (4\sqrt3 - 2\pi) = 10\pi + 4\sqrt3.$$

The answer is $\boxed{\textbf{(A)}\ 10\pi + 4\sqrt3}$.

For regions trapped between touching circles, connect the centers. The shape you get splits into the curved region plus circle slices, and each slice's angle is just the corner angle of the shape. Three equal circles that all touch each other always give three $60^\circ$ slices, which add to half a circle.

Estimate. The gap is small, so the answer should be just a little more than $12\pi \approx 37.7$. Using $\pi \approx 3.14$ and $\sqrt3 \approx 1.73$:

Only (A) is just a little above $37.7$; every other choice needs a gap of at least $1.4$. The gap is small: $4\sqrt3 - 2\pi \approx 6.93 - 6.28 = 0.65$. So the total is about $38.3$, which is (A). Estimating like this is a quick way to catch a missing or extra piece.