Example 5 Hexagon $A B C D E F$ is inscribed in $\odot O$, and $A B=B C=C D=\sqrt{3}+1, D E=E F=F A=1$. Find the area of this hexagon.

Analysis: To find the area of any hexagon, a direct approach can be quite troublesome. Observing Figure 4, we notice that it has two sets of congruent triangles, and the three triangles in each set are adjacent. Based on this characteristic of the figure, we can appropriately transform and reorganize the figure to make it easier to solve using common geometric patterns.
Solution: Connect $O A, O B, O C, O D, O E, O F$, it is easy to see that
$$
\begin{array}{l}
S_{\triangle M O B}=S_{\triangle B O C}=S_{\triangle C O D}, \\
S_{\triangle D O E}=S_{\triangle E O F}=S_{\triangle F O A} .
\end{array}
$$

Reorganize the six triangles to form a cyclic hexagon $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime}$, as shown in Figure 5. Here, $\odot O$ and $\odot O^{\prime}$ are congruent circles, and
$$
\begin{array}{l}
A^{\prime} B^{\prime}=C^{\prime} D^{\prime} \\
=E^{\prime} F^{\prime}=\sqrt{3}+1, \\
B^{\prime} C^{\prime}=D^{\prime} E^{\prime} \\
=F^{\prime} A^{\prime}=1 .
\end{array}
$$

Thus, the hexagon $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime}$ has the same area as the hexagon $A B C D E F$. To find the area of the hexagon $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime}$, extend $A^{\prime} B^{\prime}, C^{\prime} D^{\prime}, E^{\prime} F^{\prime}$ to get the intersection points $L, M, N$.

Since the six interior angles of $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime}$ are equal, each being $120^{\circ}$, $\triangle L M N$ is an equilateral triangle.
$$
\begin{array}{l}
\text { Therefore, } S_{\text {hexagon } A B C D E F}=S_{\text {hexagon } A^{\prime} B C D E F} \\
=S_{\triangle L M N}-3 S_{\triangle N A^{\prime} F^{\prime}} \\
=\frac{\sqrt{3}}{4}(3+\sqrt{3})^{2}-3 \times \frac{\sqrt{3}}{4} \times 1^{2}=\frac{9}{4}(2+\sqrt{3}) .
\end{array}
$$