Initially 243 Hexagon $A B C D E F$ is inscribed in $\odot O$, $A F // D C, F E // C B, E D // B A, A B+B C=$ $2 C D$. Constructing squares on each of the six sides, the sum of the areas of these six squares is 2008. Find the perimeter of hexagon $A B C D E F$.
Solution: As shown in Figure 3, from $A F$
$/ / D C, F E / / C B, E D / /$
$B A$, we know that $A D, B E, C F$ are
all diameters of $\odot O$, thus
$$
\begin{array}{l}
D E=A B, \\
E F=B C, \\
F A=C D .
\end{array}
$$

Let $C D=x, A B=x-d$. Then
$$
\begin{array}{l}
B C=x+d, D E=x-d, \\
A F=x, E F=x+d .
\end{array}
$$

According to the problem, we have
$$
2(x-d)^{2}+2 x^{2}+2(x+d)^{2}=2008,
$$

which simplifies to $(x-d)^{2}+x^{2}+(x+d)^{2}=1004$.
Thus, $3 x^{2}+2 d^{2}=1004$.
Clearly, $x^{2}=\frac{1004-2 d^{2}}{3}1004$, so $x^{2}>200$. Therefore, $x \geqslant 15$.
Noting that $x$ is an even number, we have $x=16$ or 18.
If $x=16$, then $d^{2}=\frac{1004-3 x^{2}}{2}=118$, which clearly has no solution.

If $x=18$, then $d^{2}=\frac{1004-3 x^{2}}{2}=16$, so $d=4$. In this case, $A B=x-d=14, C D=x=18, B C$ $=x+d=22$.
Therefore, the perimeter of the hexagon $A B C D E F$ is
$$
\begin{array}{l}
A B+B C+C D+D E+E F+F A \\
=2(14+18+22)=108 .
\end{array}
$$