[Pythagorean Theorem (direct and inverse).]

The sum of the legs of a right triangle is $l$, and the height dropped from the vertex of the right angle is $h$. Find the area of the triangle.
Let $CD = h$ be the height of the right triangle $ABC$ dropped from the vertex of the right angle $C$, and $S$ be the area of the triangle. Denote $AC = x, BC = y$. Then

\[
\begin{gathered}
AB = \sqrt{x^2 + y^2} = \sqrt{(x + y)^2 - 2xy} = \sqrt{l^2 - 4S} \\
S = \frac{1}{2} CD \cdot AB = \frac{1}{2} h \sqrt{l^2 - 4S}
\end{gathered}
\]

From the equation $\frac{1}{2} h \sqrt{l^2 - 4S} = S$, we find that $S = \frac{1}{2} h \left(\sqrt{l^2 + h^2} - h\right)$.

## Answer

$\frac{1}{2} h \left(\sqrt{l^2 + h^2} - h\right)$