19.8 For a regular triangle, a square, and a regular hexagon with the same perimeter, their areas are $S_{3}$, $S_{4}$, $S_{6}$, respectively, then
(A) $S_{3}>S_{4}>S_{6}$.
(B) $S_{6}>S_{4}>S_{3}$.
(C) $S_{6}>S_{3}>S_{4}$.
(D) $S_{3}>S_{6}>S_{4}$.
(E) $S_{4}>S_{6}>S_{3}$.
(China Junior High School Mathematics League, 1984)
[Solution] Let the same perimeter be $12 l(l>0)$, then their areas are respectively
$$
\begin{aligned}
S_{3} & =\frac{1}{2} \cdot 4 l \cdot 4 l \cdot \sin 60^{\circ}=4 \sqrt{3} l^{2}, \\
S_{4} & =3 l \cdot 3 l=9 l^{2}, \\
S_{6} & =6 \sqrt{3} l^{2}, \\
\therefore \quad S_{6}>S_{4} & >S_{3} .
\end{aligned}
$$

Therefore, the correct choice is (B).