Example 6 Given that the area of quadrilateral $ABCD$ is 32, the lengths of $AB$, $CD$, and $AC$ are all integers, and their sum is 16.
(1) How many such quadrilaterals are there?
(2) Find the minimum value of the sum of the squares of the side lengths of such quadrilaterals.
(2003, National Junior High School Mathematics League)
(1) As shown in Figure 3, let $A B=a, C D=b, A C=l$, and let the height from $A B$ in $\triangle A B C$ be $h_{1}$, and the height from $D C$ in $\triangle A D C$ be $h_{2}$.
Then
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D} \\
=S_{\triangle A B C}+S_{\triangle A D C} \\
=\frac{1}{2}\left(h_{1} a+h_{2} b\right) \\
\leqslant \frac{1}{2} l(a+b) .
\end{array}
$$

Equality holds if and only if $h_{1}=h_{2}=l$, i.e., $A C \perp A B$ and $A C \perp A D$.
From the given, we have $32 \leqslant \frac{1}{2} l(a+b)$.
Also, by the problem's condition, $a+b=16-l$, so we get
$$
64 \leqslant l(16-l)=64-(l-8)^{2} \leqslant 64 \text {. }
$$

Thus, it must be that $l=8, a+b=8$, and at this time, $A C \perp A B$ and $A C \perp C D$.
Therefore, such quadrilaterals are as follows:
$$
\begin{array}{l}
a=1, b=7, l=8 ; a=2, b=6, l=8 ; \\
a=3, b=5, l=8 ; a=b=4, l=8 .
\end{array}
$$

These are all trapezoids or parallelograms with $A C$ as the height.
(2) From $A B=a, C D=8-a$, we get
$$
B C^{2}=8^{2}+a^{2}, A D^{2}=8^{2}+(8-a)^{2} \text {. }
$$

Therefore, the sum of the squares of the side lengths of such quadrilaterals is
$$
\begin{array}{l}
2 a^{2}+2(8-a)^{2}+2 \times 8^{2} \\
=4(a-4)^{2}+192 .
\end{array}
$$

Hence, when $a=b=4$, the above sum of squares is minimized, and the minimum value is 192.