$19 \cdot 10$ A regular hexagon $A B C D E F$ has a smaller regular hexagon formed by connecting the midpoints of its sides. The area of the smaller hexagon is what fraction of the area of the larger hexagon?
(A) $\frac{1}{2}$.
(B) $\frac{\sqrt{3}}{3}$.
(C) $\frac{2}{3}$.
(D) $\frac{3}{4}$.
(E) $\frac{\sqrt{3}}{2}$.
(47th American High School Mathematics Examination, 1996)
［Solution］The ratio of the areas of similar figures is equal to the square of the ratio of their corresponding side lengths.
Now consider the corner of the large regular hexagon with vertex $A$ that is cut off by the small regular hexagon (as shown in the figure).
Obviously, $\angle M A N=120^{\circ}$, let $A M=a$.
Draw $A H \perp M N$ at $H$, then $M H=\frac{\sqrt{3}}{2} a$.
$$
\therefore \quad \frac{S_{\text {large regular hexagon }}}{S_{\text {small regular hexagon }}}=\frac{(2 A M)^{2}}{(2 A H)^{2}}=\left(\frac{A M}{A H}\right)^{2}=\frac{3}{4} \text {. }
$$

Therefore, the answer is $(D)$.