Example 5 Does a right-angled triangle with a perimeter of 6 and an integer area exist? If not, please provide a proof; if it exists, please specify how many there are?
(1904, National Junior High School Mathematics League)
Solution: Such a right-angled triangle exists, and there is exactly one.

Let the hypotenuse of this right-angled triangle be $c$, the two legs be $a, b (a \leqslant b)$, and the area be $S$. Then
$$
\begin{array}{l}
a+b+c=6, a^{2}+b^{2}=c^{2} . \\
\therefore a+b=6-c, a b=18-6 c .
\end{array}
$$

It is evident that $a, b$ are the roots of the equation
$$
x^{2}-(6-c) x+(18-6 c)=0
$$

Thus,
$$
\Delta=(6-c)^{2}-4(18-6 c) \geqslant 0 .
$$

Solving this, we get $c \geqslant 6 \sqrt{2}-6 \approx 2.4$.
From $c<a+b=6-c$ we get $c<3$.
$\because S=\frac{1}{2} a b=9-3 c$ is an integer,
$\therefore 3 c$ is an integer.
Hence, from $7.2 \leqslant 3 c<9$ we know $3 c=8$, i.e., $c=\frac{8}{3}$.
It is easy to find $a=\frac{5-\sqrt{7}}{3}, b=\frac{5+\sqrt{7}}{3}$.
Thus, we obtain the unique right-angled triangle.