Set the sum of the three side lengths of $\triangle ABC$ equal to the given perimeter. Then use the matching tick marks.

In $\triangle ADE$, the angle at $A$ is still $30^\circ$. Relate $DE$ to $AD$ with tangent, then use the triangle-area formula.

Algebra is faster for this problem. For a Desmos verification, graph y=x(3+sqrt(3)) and y=24+8sqrt(3). Click their intersection to confirm that $x=8$.

Enter (1/2)(8)(8/sqrt(3)). Desmos displays the decimal value of the shaded area; the exact form is found by rationalizing the denominator.

Let $BC=x$. Since $\angle A=30^\circ$ and $\angle B=90^\circ$, $\triangle ABC$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle. Its side lengths are $x$, $x\sqrt{3}$, and $2x$.

Therefore, its perimeter is

Set this equal to the given perimeter:

so $x=8$. Thus, $BC=8$.

The matching tick marks show that $AD=BC$. Therefore, $AD=8$.

Triangle $ADE$ is a right triangle with $\angle A=30^\circ$. Using $\tan 30^\circ=\frac{1}{\sqrt{3}}$,

Thus,

The shaded region is $\triangle ADE$, so