10.341. The hypotenuse of a right triangle is equal to $c$. The projection of the vertex of the right angle onto the hypotenuse divides it into two segments, of which the smaller is to the larger as the larger is to the entire hypotenuse. Determine the area of the triangle.
Solution.

Let $x$ be the larger segment of the hypotenuse. Then, by the condition, $\frac{c-x}{x}=\frac{x}{c}$, $x^{2}+c x-c^{2}=0$, from which $x=\frac{c(\sqrt{5}-1)}{2}$ (the second root of the equation does not fit). Hence, $x^{2}=\frac{c^{2}(3-\sqrt{5})}{2}$. Denoting the height to the hypotenuse by $h$, we have $\frac{c-x}{h}=\frac{h}{x}$, and thus, $h^{2}=c x-x^{2}=$ $=c^{2}\left(\frac{\sqrt{5}-1}{2}-\frac{3-\sqrt{5}}{2}\right)=c^{2}(\sqrt{5}-2)$, i.e., $h=c \sqrt{\sqrt{5}-2}$. Therefore, $S=\frac{1}{2} c h=\frac{c^{2} \sqrt{\sqrt{5}-2}}{2}$.

Answer: $0.5 c^{2} \sqrt{\sqrt{5}-2}$.