2. Positive numbers $a, b, c$ satisfy $\left\{\begin{array}{l}a+b+c=10, \\ a^{2}+b^{2}=c^{2} .\end{array}\right.$ Then the maximum value of $ab$ is $\qquad$
2. $50(3-2 \sqrt{2})$.

With sides $a, b, c$, a right triangle $\triangle ABC$ can be constructed, with a perimeter of 10. Since among right triangles with a fixed perimeter, the isosceles right triangle has the maximum area, when
$$
a=b=5(2-\sqrt{2}), c=10(\sqrt{2}-1)
$$

$a b$ has the maximum value of $50(3-2 \sqrt{2})$.