jmc

Recognizing that all our triangles in the diagram are 30-60-90 triangles, we recall that the ratio of the longer leg to the hypotenuse in such a triangle is $\frac{\sqrt{3}}{2}$. Therefore, we can see that: $$\begin{array}{rl}AB& =24\left(\frac{\sqrt{3}}{2}\right)=12\sqrt{3}\\ BC& =12\left(\frac{\sqrt{3}}{2}\right)=6\sqrt{3}\\ CD& =6\left(\frac{\sqrt{3}}{2}\right)=3\sqrt{3}\\ ED& =6\left(\frac{1}{2}\right)=3\end{array}$$ The perimeter of quadrilateral $ABCD$ is equal to $AB+BC+CD+DA$ and $DA=DE+EA$, so the perimeter is $12\sqrt{3}+6\sqrt{3}+3\sqrt{3}+3+24=\boxed{27+21\sqrt{3}}$.