10.416 Determine the area of a triangle given its three heights $h_{1}$, $h_{2}, h_{3} \cdot$
Solution. Let $a, b, c$ be the sides of the given triangle, to which the heights $h_{1}, h_{2}, h_{3}$ are drawn, respectively. Then $a=\frac{2 S}{h_{1}}, b=\frac{2 S}{h_{2}}, c=\frac{2 S}{h_{3}}$, where $S$ is the area of the given triangle. Substituting these expressions into Heron's formula, we have:

$$
S=\sqrt{\left(\frac{s}{h_{1}}+\frac{s}{h_{2}}+\frac{s}{h_{3}}\right)\left(\frac{s}{h_{1}}+\frac{s}{h_{2}}-\frac{s}{h_{3}}\right)\left(\frac{s}{h_{1}}+\frac{s}{h_{3}}-\frac{s}{h_{2}}\right)\left(\frac{S}{h_{2}}+\frac{S}{h_{3}}-\frac{s}{h_{1}}\right)}
$$

From this, we have

Answer: $s=$

$$
\sqrt{\left(\frac{1}{h_{1}}+\frac{1}{h_{2}}+\frac{1}{h_{3}}\right)\left(\frac{1}{h_{1}}+\frac{1}{h_{2}}-\frac{1}{h_{3}}\right)\left(\frac{1}{h_{1}}+\frac{1}{h_{3}}-\frac{1}{h_{2}}\right)\left(\frac{1}{h_{2}}+\frac{1}{h_{3}}-\frac{1}{h_{1}}\right)}
$$