For example, the area of $\triangle ABC$ is three times the area of its inscribed rectangle $PQRS$, and the values of side $BC$ and altitude $AU$ are both rational numbers. Under what conditions is the perimeter of rectangle $PQRS$ a rational number? Under what conditions is it an irrational number?
(1985, Guangzhou and Four Other Cities Junior High School Mathematics League)
Solution: As shown in Figure 2, let
$$
\begin{array}{c}
BC=a, AD=h, \\
PQ=x, PS=y. \text{ Then } \\
\frac{h-y}{h}=\frac{x}{a},
\end{array}
$$

Also, $xy=\frac{1}{3}$
$$
\cdot \frac{1}{2} a h=\frac{1}{6} a h \text{. }
$$

From (1), let $ay=\frac{ah}{2}+t, hx=\frac{ah}{2}-t$. Then
$$
x=\frac{a}{2}-\frac{t}{h}, y=\frac{h}{2}+\frac{t}{a} \text{. }
$$

Substituting into (2) yields
$$
\left(\frac{a}{2}-\frac{t}{h}\right)\left(\frac{h}{2}+\frac{t}{a}\right)=\frac{1}{6} a h .
$$

Rearranging and simplifying gives $t^{2}=\frac{1}{12} a^{2} h^{2}$.
$$
\therefore t= \pm \frac{\sqrt{3}}{6} a h \text{. }
$$

Therefore, the perimeter of rectangle $PQRS$ is
$$
\begin{array}{l}
2(x+y)=2\left(\left(\frac{a}{2}-\frac{t}{h}\right)+\left(\frac{h}{2}+\frac{t}{a}\right)\right) \\
=2\left(\frac{a+h}{2}+\frac{h-a}{ah} t\right) \\
=a+h \pm \frac{\sqrt{3}}{3}(h-a) .
\end{array}
$$

Since $a$ and $h$ are both rational numbers,
when $a=h$, i.e., $BC=AD$, the perimeter of rectangle $PQRS$ is a rational number;

when $a \neq h$, i.e., $BC \neq AD$, the perimeter of rectangle $PQRS$ is an irrational number.