jmc

All of our triangles in this diagram are 30-60-90 triangles. We know that the ratio of the side lengths in a 30-60-90 triangle is $1:\sqrt{3}:2.$

Since $AE=24$ and $\angle AEB=60^{\circ }$ and $AEB$ is a right triangle, then we can see that $AE$ is the hypotenuse and $BE$ is the shorter leg, so $BE=\frac{1}{2}\cdot 24=12.$ Likewise, since $BE=12$ and $\angle BEC=60^{\circ }$, then $CE=\frac{1}{2}\cdot 12=6$. Then, $AB=24\left(\frac{\sqrt{3}}{2}\right)=12\sqrt{3}$ and $BC=12\left(\frac{\sqrt{3}}{2}\right)=6\sqrt{3}.$ Continuing, we find that $CD=6\left(\frac{\sqrt{3}}{2}\right)=3\sqrt{3}$ and $ED=6\left(\frac{1}{2}\right)=3.$

The area of quadrilateral $ABCD$ is equal to the sum of the areas of triangles $ABE$, $BCE$ and $CDE$. Thus, $$\begin{array}{rl}\text{mbox}Area& =\frac{1}{2}(BE)(BA)+\frac{1}{2}(CE)(BC)+\frac{1}{2}(DE)(DC)\\ & =\frac{1}{2}(12)(12\sqrt{3})+\frac{1}{2}(6)(6\sqrt{3})+\frac{1}{2}(3)(3\sqrt{3})\\ & =72\sqrt{3}+18\sqrt{3}+\frac{9}{2}\sqrt{3}\\ & =\boxed{\frac{189}{2}\sqrt{3}}\end{array}$$