\[\boxed{\mathbf{1221.ОК\ ГДЗ - домашка\ н}а\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABC - конус;\]
\[S_{осн} = Q;\]
\[S_{бок} = P.\]
\[\mathbf{Найти:}\]
\[V - ?\]
\[\mathbf{Решение.}\]
\[1)\ V = \frac{1}{3}\pi r^{2}h;\ \ \ \]
\[r - радиус\ основания;\ \ \]
\[h - высота.\]
\[2)\ S_{осн} = Q\ (по\ условию);\ \ \]
\[S_{осн} = \pi r^{2}:\]
\[\pi r^{2} = Q\]
\[r^{2} = \frac{Q}{\pi}\]
\[r = \sqrt{\frac{Q}{\pi}}.\]
\[3)\ S_{бок} = \pi rl;\ \ \ \]
\[l = AB - образующая\ конуса.\]
\[S_{бок} = P \Longrightarrow \pi rl = P:\]
\[l = \frac{P}{\text{πr}} = \frac{P}{\pi\sqrt{\frac{Q}{\pi}}} = \frac{P}{\sqrt{\text{πQ}}}.\]
\[4)\ \mathrm{\Delta}AOB - прямоугольный:\]
\[h = \sqrt{l^{2} - r^{2}} = \sqrt{\frac{P^{2}}{\text{πQ}} - \frac{Q}{\pi}} =\]
\[= \sqrt{\frac{P^{2} - Q^{2}}{\text{πQ}}}.\]
\[5)\ V = \frac{1}{3}\pi \bullet \frac{Q}{\pi} \bullet \sqrt{\frac{P^{2} - Q^{2}}{\text{πQ}}} =\]
\[= \frac{1}{3}Q\sqrt{\frac{P^{2} - Q^{2}}{\text{πQ}}} =\]
\[= \frac{1}{3}\sqrt{\frac{Q\left( P^{2} - Q^{2} \right)}{\pi}}.\]
\[\mathbf{Ответ:\ }\frac{1}{3}\sqrt{\frac{Q\left( P^{2} - Q^{2} \right)}{\pi}}\mathbf{.}\]