2014 AMC 10A Problem 12

Hexagon. A regular hexagon of side $6$ splits into six equilateral triangles of side $6$, each of area $\frac{\sqrt3}{4}\cdot 36 = 9\sqrt3$. Total: $54\sqrt3$.

Sectors. At each vertex the sector is bounded by the two sides meeting there, so its angle is the interior angle of the hexagon, $120^\circ$, one third of a circle. Each sector has area $\frac13 \cdot \pi \cdot 3^2 = 3\pi$. Six of them give $18\pi$. (Since the radius $3$ is half the side, neighboring sectors just touch at the midpoints of the sides and do not overlap.)

Shaded region. Subtract:
$$54\sqrt3 - 18\pi.$$

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

"Inside the polygon, outside the sectors" is a subtraction of areas, and the only thing to get right is each sector's angle, which equals the polygon's interior angle at that vertex. A useful shortcut: the interior angles of any $n$-gon total $(n-2)\cdot 180^\circ$, so equal-radius sectors at every vertex always add up to $\frac{n-2}{2}$ full circles, here $2$ circles of radius $3$.

• Using 60 degrees for each sector (the central angle of a hexagon) instead of the 120-degree interior angle, giving 54 root 3 - 9 pi.
• Computing the hexagon area with side $3$ or as $\frac{3\sqrt3}{2}s^2$ with an arithmetic slip, landing on $27\sqrt3$.