Example 5 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)
Analysis: Note that the number of quadrilaterals $ABCD$ is determined by the number of triples of the lengths of $AB$, $CD$, and $AC$. Therefore, we can start by determining the shape of the quadrilateral $ABCD$.

Solution: (1) As shown in Figure 2, let
$$
AB = a, CD = b, AC = l
$$
$(a, b, l$ are all positive integers, and $a \leq b$). Let the height from $AB$ in $\triangle ABC$ be $h_a$, and the height from $DC$ in $\triangle ACD$ be $h_b$. Then
$$
\begin{array}{l}
S_{\text{quadrilateral } ABCD} = S_{\triangle ABC} + S_{\triangle ACD} \\
= \frac{1}{2} a h_a + \frac{1}{2} b h_b \leq \frac{1}{2} (a + b) l.
\end{array}
$$

Equality holds if and only if $h_a = h_b = l$. At this point, $AB \parallel CD$, meaning that quadrilateral $ABCD$ is a trapezoid or a parallelogram, and $AC$ is their height.
From the given information,
$$
64 \leq (a + b) l.
$$

Also, from the problem statement,
$$
a + b = 16 - l.
$$

Substituting equation (2) into equation (1) gives $(16 - l) l \geq 64$, which simplifies to
$$
(l - 8)^2 \leq 0.
$$

Thus, $l = 8$. At this point, $a + b = 8$.
Therefore, there are four such quadrilaterals:
$$
\begin{array}{l}
a = 1, b = 7, l = 8; a = 2, b = 6, l = 8; \\
a = 3, b = 5, l = 8; a = 4, b = 4, l = 8.
\end{array}
$$
(2) In quadrilateral $ABCD$, $AC \perp AB$, $AC \perp CD$. If $AB = a$, from the conclusion in (1),
$$
\begin{array}{l}
CD = 8 - a, BC^2 = AB^2 + AC^2 = a^2 + 8^2, \\
AD^2 = CD^2 + AC^2 = (8 - a)^2 + 8^2,
\end{array}
$$

where $a$ is a positive integer, and $1 \leq a \leq 4$.
Therefore, the sum of the squares of the side lengths of such a quadrilateral is
$$
\begin{array}{l}
AB^2 + BC^2 + CD^2 + DA^2 \\
= a^2 + (a^2 + 8^2) + (8 - a)^2 + [(8 - a)^2 + 8^2] \\
= 4a^2 - 32a + 256 = 4(a - 4)^2 + 192 \geq 192.
\end{array}
$$

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