II. (15 points) Inside a rectangle $ABCD$ (excluding the boundary), there is a point $P$, which is 1 unit away from vertex $A$ and sides $BC$, $CD$. Find the range of possible areas for rectangle $ABCD$.
II. As shown in Figure 10, $PE \perp BC$ at $E$, $PF \perp CD$ at $F$, then $PE = PF = 1$.

Let $BE = x$, $DF = y$. Then
$$
x^{2} + y^{2} = PA^{2} = 1,
$$

and $x + y > PA = 1$.
Thus, $S_{\text{quadrilateral } ABCD} = (1 + x)(1 + y)$
$$
= 1 + x + y + xy > 2,
$$

and when $x$ (or $y) \rightarrow 0$, $S \rightarrow 2$.
Since $(x + y)^{2} \leqslant 2(x^{2} + y^{2}) = 2$,
$x + y \leqslant \sqrt{2}$, and $xy \leqslant \frac{x^{2} + y^{2}}{2} = \frac{1}{2}$,

therefore, $S_{\text{quadrilateral } ABCD} \leqslant \frac{3}{2} + \sqrt{2}$.
The equality holds if and only if $x = y = \frac{\sqrt{2}}{2}$.
Thus, the range of $S_{\text{quadrilateral } ABCD}$ is
$$
2 < S_{\text{quadrilateral } ABCD} \leqslant \frac{3}{2} + \sqrt{2}.
$$