2014 AMC 10

We note that $\triangle ADE$ is a right triangle with $\angle DAE = 30^\circ$ . Therefore, $AE=\cos 30^\circ = \dfrac {\sqrt 3}2$ and $AC = 4 \times \dfrac {\sqrt 3}2 = 2\sqrt 3$ . The area of $\triangle ABC$ is given by:

$\begin{aligned} [ABC] & = \frac 12 AB \times AC \times \sin \angle A \\ & = \frac 12 \times 2\sqrt 3 \times 2\sqrt 3 \times \frac {\sqrt 3}2 \\ & = \boxed{3\sqrt 3} \end{aligned}$