10.262. In a right-angled triangle, the height drawn to the hypotenuse is $h$; the radius of the inscribed circle is $r$. Find the hypotenuse.
Solution.

Let $a$ and $b$ be the legs, $c$ be the hypotenuse, and $p$ be the semiperimeter of the given triangle. $r=\frac{a+b-c}{2}=\frac{a+b+c}{2}-c=p-c ; p=r+c$. The area of the triangle $S=\frac{1}{2} c h=p r$. Then $\frac{1}{2} c h=(r+c) r ; c(h-2 r)=2 r^{2} ; c=\frac{2 r^{2}}{h-2 r}$. Answer: $\frac{2 r^{2}}{h-2 r}$.