10. (20 points) The lengths of two adjacent sides of a rectangle are $a$ and $b$ $(a \leqslant b)$. A straight line cuts the rectangle to form a right-angled triangle with a perimeter of $l$. Find the minimum value of the remaining area of the rectangle.
10. Let the segments intercepted by the line on the shorter and longer sides of the rectangle be $x$ and $y$, respectively. Then,
$$
\begin{array}{l}
x+y+\sqrt{x^{2}+y^{2}}=l \\
\Rightarrow \sqrt{x^{2}+y^{2}}=l-x-y \\
\Rightarrow(l-x)(l-y)=\frac{l^{2}}{2} .
\end{array}
$$

First, find the maximum area of the right triangle intercepted:
$$
S=\frac{1}{2} x y=-\frac{l}{2}\left[(l-x)+\frac{l^{2}}{2(l-x)}\right]+\frac{3 l^{2}}{4},
$$

where, $l-a \leqslant l-x\frac{l}{\sqrt{2}}$, i.e., $a<\left(1-\frac{\sqrt{2}}{2}\right) l$, then,
$$
S \leqslant \frac{a l(l-2 a)}{4(l-a)},
$$

with equality holding if and only if $x=a$.
Thus, the minimum area of the remaining part of the rectangle is
$$
\begin{array}{l}
a b-S_{\max } \\
=\left\{\begin{array}{ll}
a b-\left(\frac{3}{4}-\frac{\sqrt{2}}{2}\right) l^{2}, & a \geqslant\left(1-\frac{\sqrt{2}}{2}\right) l \\
a b-\frac{a l(l-2 a)}{4(l-a)}, & a<\left(1-\frac{\sqrt{2}}{2}\right) l
\end{array}\right.
\end{array}
$$