$19 \cdot 9$ If $P_{1} P_{2} P_{3} P_{4} P_{5} P_{6}$ is a regular hexagon. Its apothem (the distance from the center to the midpoint of a side) is 2, and $Q_{i}$ is the midpoint of side $P_{i} P_{i+1}$, where $i=1,2,3,4$, then the area of quadrilateral $Q_{1} Q_{2} Q_{3} Q_{4}$ is
(A) 6.
(B) $2 \sqrt{6}$.
(C) $\frac{8 \sqrt{3}}{3}$.
(D) $3 \sqrt{3}$.
(30th American High School Mathematics Examination, 1979)
[Solution] If $O$ is the center of the hexagon, then $O Q_{i}(i=1,2,3,4)$ are the apothems of the regular hexagon.

Thus, the area of $Q_{1} Q_{2} Q_{3} Q_{4}$ is the sum of the areas of three equilateral triangles $\triangle Q_{1} Q_{2} O, \triangle Q_{2} Q_{3} O, \triangle Q_{3} Q_{4} O$, each with a side length of 2.
$$
\text { Therefore } \begin{aligned}
S_{Q_{1} Q_{2} Q_{3} Q_{4}} & =3\left(\frac{2^{2} \cdot \sqrt{3}}{4}\right) \\
& =3 \sqrt{3} .
\end{aligned}
$$

Hence, the answer is $(D)$.