11.054. Each of the lateral edges of the pyramid is equal to $b$. Its base is a right triangle, the legs of which are in the ratio $m: n$, and the hypotenuse is equal to $c$. Calculate the volume of the pyramid.
Solution.

$V=\frac{1}{3} S \cdot H$. Let $A B=m x, B C=n x$ (Fig. 11.49). By the condition $A C=c, S C=b$. In $\triangle A B C, \angle A B C=90^{\circ}$; thus, $c^{2}=m^{2} x^{2}+n^{2} x^{2}$, $x=\sqrt{\frac{c^{2}}{m^{2}+n^{2}}}=\frac{c}{\sqrt{m^{2}+n^{2}}}$ and $S=\frac{1}{2} m n x^{2}=\frac{m n c^{2}}{2\left(m^{2}+n^{2}\right)}$. Since

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-683.jpg?height=596&width=420&top_left_y=92&top_left_x=124)

Fig. 11.49

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-683.jpg?height=560&width=566&top_left_y=94&top_left_x=681)

Fig. 11.50

$H=S O$ and $A O=O C=B O=\frac{c}{2}$, then $H^{2}=b^{2}-\frac{c^{2}}{4} ; H=\frac{\sqrt{4 b^{2}-c^{2}}}{2}$ (in $\triangle S O C$ $\left.\angle S O C=90^{\circ}\right) ; V=\frac{1}{3} \frac{m n}{2} \cdot \frac{c^{2} \sqrt{4 b^{2}-c^{2}}}{\left(m^{2}+n^{2}\right) \cdot 2}=\frac{m n c^{2} \sqrt{4 b^{2}-c^{2}}}{12\left(m^{2}+n^{2}\right)}$.

Answer. $\frac{m n c^{2} \sqrt{4 b^{2}-c^{2}}}{12\left(m^{2}+n^{2}\right)}$.