imc

We note that the $6$ triangular sections in $△ABC$ 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 $\frac{3\sqrt{3}}{2}$. The area of $△ABC$, which is equivalent to two of these hexagons together, is $\boxed{3\sqrt{3}}$.