Example 6 In the interior of rectangle $ABCD$ (excluding the boundary), there is a point $P$, which is at a distance of 1 from vertex $A$ and sides $BC$, $CD$. Find the range of the area of rectangle $ABCD$.
$(2008$, New Knowledge Cup Shanghai Junior High School Mathematics Competition)
Solve as shown in Figure 10, introduce variables $x, y$.
Then $x^{2}+y^{2}=P A^{2}=1$, and $x+y>P A=1$.
Thus, $S_{\text {rectangle } A B C D}=(1+x)(1+y)$
$$
=1+x+y+x y>2 \text {, }
$$

and when $x$ (or $y) \rightarrow 0$, $S \rightarrow 2$.
Also, $(x+y)^{2} \leqslant 2\left(x^{2}+y^{2}\right)=2$, then $x+y \leqslant \sqrt{2}$,

and $x y \leqslant \frac{x^{2}+y^{2}}{2}=\frac{1}{2}$.
Therefore, $S_{\text {rectangle } A B C D} \leqslant \frac{3}{2}+\sqrt{2}$.
The equality holds if and only if $x=y=\frac{\sqrt{2}}{2}$.
Thus, $2<S_{\text {rectangle } A B C D} \leqslant \frac{3}{2}+\sqrt{2}$.