$21 \cdot 2$ Two congruent $30^{\circ}-60^{\circ}-90^{\circ}$ triangles are placed so that part of them overlap and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area of the common part of the two triangles is 
(A) $6 \sqrt{3}$.
(B) $8 \sqrt{3}$.
(C) $9 \sqrt{3}$.
(D) $12 \sqrt{3}$.
(E) 24.
(24th American High School Mathematics Examination, 1973)
[Solution] As shown in the figure, let $M$ and $N$ be the endpoints of the height of $\triangle A B N$ on the base $A B$, which is also the median of the isosceles triangle that is the common part of the two triangles.
Notice that $\angle N A B=30^{\circ}$, then in the right triangle $\triangle A N M$,

that is
$$
\begin{array}{l}
h \sqrt{3}=6, \\
h=2 \sqrt{3} .
\end{array}
$$

The area of the common part is
$$
\frac{1}{2} h(2 A M)=\frac{1}{2}(2 \sqrt{3}) \cdot 12=12 \sqrt{3} \text {. }
$$

Therefore, the answer is $(D)$.