[ Mean Proportional in a Right Triangle ]

The height of a right triangle, dropped from the vertex of the right angle, is $h$, and the difference between the projections of the legs on the hypotenuse is $l$. Find the area of this triangle.
Let $CD = h$ be the height of the right triangle $ABC$ dropped from the vertex of the right angle $C$.

Assume that $AD \leqslant BD$. Let $AD = x$. Then,

$$
x = \frac{-l + \sqrt{l^2 + 4h^2}}{2}, \quad AB = 2x + l = \sqrt{l^2 + 4h^2}
$$

Therefore,

$$
S_{\triangle ABC} = \frac{1}{2} CD \cdot AB = \frac{1}{2} h \sqrt{l^2 + 4h^2}
$$

## Answer

$\frac{1}{2} h \sqrt{l^2 + 4h^2}$.