2014 AMC 10A Problem 12

The shaded part is the hexagon with the six sectors taken out, so find the two areas and subtract.

The hexagon.

1. A regular hexagon with side $6$ splits into six equilateral triangles with side $6$ by drawing lines from the center to each corner.
2. An equilateral triangle with side $s$ has area $\frac{\sqrt3}{4}s^2$. With $s = 6$ that is $\frac{\sqrt3}{4} \cdot 36 = 9\sqrt3$.
3. Six of them: $6 \cdot 9\sqrt3 = 54\sqrt3$.

The sectors.

1. Each sector sits in a corner of the hexagon, between the two sides that meet there. So it opens exactly as wide as the corner angle.
2. The corner angles of a hexagon add up to $(6 - 2) \cdot 180^\circ = 720^\circ$, so each corner is $720^\circ \div 6 = 120^\circ$. That is $\frac{120}{360} = \frac13$ of a full circle.
3. A full circle of radius $3$ has area $9\pi$, so each sector has area $\frac13 \cdot 9\pi = 3\pi$. Six sectors: $18\pi$.
4. The sectors do not overlap: the radius $3$ is half of a side, so two neighbouring sectors just meet at the middle of the side between them.

Shaded area. $54\sqrt3 - 18\pi$.

The answer is $\boxed{\textbf{(C)}\ 54\sqrt{3}-18\pi}$.

"Inside the shape but outside the sectors" means subtract. The only thing to be careful about is the angle of each sector, which is the polygon's corner angle, not the angle at the center. A quick check: the six corner angles add to $720^\circ$, which is exactly two full circles, so the sectors together make two circles of radius $3$, area $2 \cdot 9\pi = 18\pi$.