ГДЗ по геометрии 8 класс Атанасян Задание 1221

\[\boxed{\mathbf{1221.ОК\ ГДЗ - домашка\ н}а\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABC - конус;\]

\[S_{осн} = Q;\]

\[S_{бок} = P.\]

\[\mathbf{Найти:}\]

\[V - ?\]

\[\mathbf{Решение.}\]

\[1)\ V = \frac{1}{3}\pi r^{2}h;\ \ \ \]

\[r - радиус\ основания;\ \ \]

\[h - высота.\]

\[2)\ S_{осн} = Q\ (по\ условию);\ \ \]

\[S_{осн} = \pi r^{2}:\]

\[\pi r^{2} = Q\]

\[r^{2} = \frac{Q}{\pi}\]

\[r = \sqrt{\frac{Q}{\pi}}.\]

\[3)\ S_{бок} = \pi rl;\ \ \ \]

\[l = AB - образующая\ конуса.\]

\[S_{бок} = P \Longrightarrow \pi rl = P:\]

\[l = \frac{P}{\text{πr}} = \frac{P}{\pi\sqrt{\frac{Q}{\pi}}} = \frac{P}{\sqrt{\text{πQ}}}.\]

\[4)\ \mathrm{\Delta}AOB - прямоугольный:\]

\[h = \sqrt{l^{2} - r^{2}} = \sqrt{\frac{P^{2}}{\text{πQ}} - \frac{Q}{\pi}} =\]

\[= \sqrt{\frac{P^{2} - Q^{2}}{\text{πQ}}}.\]

\[5)\ V = \frac{1}{3}\pi \bullet \frac{Q}{\pi} \bullet \sqrt{\frac{P^{2} - Q^{2}}{\text{πQ}}} =\]

\[= \frac{1}{3}Q\sqrt{\frac{P^{2} - Q^{2}}{\text{πQ}}} =\]

\[= \frac{1}{3}\sqrt{\frac{Q\left( P^{2} - Q^{2} \right)}{\pi}}.\]

\[\mathbf{Ответ:\ }\frac{1}{3}\sqrt{\frac{Q\left( P^{2} - Q^{2} \right)}{\pi}}\mathbf{.}\]