$[\quad$ Pyramid (other). $\quad]$

The base of a triangular pyramid is a right triangle with legs $a$ and $b$. The lateral edges are equal to $l$. Find the height of the pyramid.
Let $S A B C$ be the given pyramid. Since its lateral edges are equal, the height $S O$ passes through the center $O$ of the circle circumscribed around the base. Since the base of the pyramid is a right triangle $A B C$, $O$ is the midpoint of its hypotenuse $A B$, and the radius of the circumscribed circle of the triangle is half the hypotenuse, i.e.,

$$
R=\frac{1}{2} \sqrt{a^{2}+b^{2}}
$$

From the right triangle $A O S$, by the Pythagorean theorem, we find that

$$
S O=\sqrt{S A^{2}-O A^{2}}=\sqrt{l^{2}-\frac{1}{4}\left(a^{2}+b^{2}\right)}=\frac{1}{2} \sqrt{4 l^{2}-a^{2}-b^{2}}
$$

## Problem