Four. (This question is worth 20 points) In the cyclic hexagon $A B C D E F$, $A B=B C=C D=3 \text{~cm}, D E=E F=$ $F A=5 \text{~cm}$. Find the area of this hexagon.

.
$$
\begin{array}{l}
=\frac{1}{2}(\overparen{B C}+\overparen{A B}+\overparen{A F}+\overparen{E F}) \\
=\frac{1}{2} \cdot \frac{2}{3} \cdot 360^{\circ}=120^{\circ} .
\end{array}
$$

Similarly, $\angle B A F=120^{\circ}$.
By the cosine rule
$$
\begin{array}{l}
C E^{2}=C D^{2}+D E^{2}-2 C D \cdot D E \cos 120^{\circ} \\
=3^{2}+5^{2}-2 \cdot 3 \cdot 5\left(-\frac{1}{2}\right)=49 . \\
\therefore C E=7 .
\end{array}
$$

Similarly, $B F=7$.
$$
\begin{array}{l}
\because C E=B F, \\
\therefore B C / / E F .
\end{array}
$$
$\therefore$ Quadrilateral $B C E F$ is an isosceles trapezoid.
Draw $B G \perp E F$ and
intersect $E F$ at $G$, then
$$
\begin{array}{l}
\quad G F=\frac{1}{2}(E F \\
-B C)=\frac{1}{2}(5-3) \\
=1 .
\end{array}
$$

In the right triangle $\triangle B G F$, $G F=1, E F=7$, thus, $B G=\sqrt{13 F^{2}-C F^{2}}=\sqrt{7^{2}-1^{2}}=4 \sqrt{3}$.
$$
\begin{aligned}
S_{\triangle C D E} & =\frac{1}{2} C D \cdot D E \sin \angle C D E \\
& =\frac{1}{2} \cdot 3 \cdot 5 \cdot \frac{\sqrt{3}}{2}=\frac{15}{4} \sqrt{3} .
\end{aligned}
$$

Similarly, $S_{\triangle A B F}=\frac{5}{4} \sqrt{3}$.
$$
\begin{aligned}
S_{\text {trapezoid } A B E F} & =\frac{1}{2}(B C+E F) \cdot B G \\
& =\frac{1}{2}(3+5) \cdot 4 \sqrt{3}=16 \sqrt{3} .
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
S_{\text {hexagon } A B C D E F} & =S_{\triangle C D E}+S_{\triangle A B F}+S_{\text {trapezoid } A B E F} \\
& =\frac{15}{4} \sqrt{3}+\frac{5}{4} \sqrt{3}+16 \sqrt{3} \\
& =\frac{47}{2} \sqrt{3}\left(\mathrm{~cm}^{2}\right) .
\end{aligned}
$$