2014 AMC 10B Problem 13

Follow the $60^\circ$ grid formed by the unit hexagons and place $A=(0,0)$. The marked vertices may then be written as $B=(3,\sqrt3)$ and $C=(3,-\sqrt3)$. Thus $BC=2\sqrt3$, and $$\begin{aligned} AB=AC&=\sqrt{3^2+(\sqrt3)^2}\\ &=2\sqrt3. \end{aligned}$$ Hence $\triangle ABC$ is equilateral, with area $$\frac{(2\sqrt3)^2\sqrt3}{4}=3\sqrt3.$$ Thus, the correct answer is B .