Six regular hexagons surround a regular hexagon of side length $1$ as shown. What is the area of $\triangle{ABC}$?


$\textbf {(A) } 2\sqrt{3} \qquad \textbf {(B) } 3\sqrt{3} \qquad \textbf {(C) } 1+3\sqrt{2} \qquad \textbf {(D) } 2+2\sqrt{3} \qquad \textbf {(E) } 3+2\sqrt{3}$
We note that the $6$ triangular sections in $\triangle{ABC}$ can be put together to form a hexagon congruent to each of the seven other hexagons. By the formula for the area of the hexagon, we get the area for each hexagon as $\dfrac{3\sqrt{3}}{2}$. The area of $\triangle{ABC}$, which is equivalent to two of these hexagons together, is $\boxed{\textbf{(B)}  3\sqrt{3}}$.