\[\boxed{\mathbf{1222.ОК\ ГДЗ - домашка\ н}а\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[конус;\]
\[S = 45\pi\ дм^{2};\]
\[\beta = 60{^\circ}.\]
\[\mathbf{Найти:}\]
\[V - ?\]
\[\mathbf{Решение.}\]
\[1)\ Длина\ окружности:\]
\[C = 2\pi r = \frac{60{^\circ}}{360{^\circ}} \bullet 2\pi l;\ \ \]
\[l = 6r.\]
\[2)\ Площадь\ боковой\ \]
\[поверхности:\ \]
\[S_{бок} = \frac{60{^\circ}}{360{^\circ}} \bullet \pi l^{2} = \frac{\pi l^{2}}{6} =\]
\[= \frac{\pi}{6}(6r)^{2} = 6\pi r^{2}\ дм^{2}.\]
\[3)\ S = S_{осн} + S_{бок} = \pi r^{2} + S_{бок} =\]
\[= \pi r^{2} + 6\pi r^{2} = 7\pi r^{2} = 45\pi\]
\[r^{2} = \frac{45}{7}.\]
\[4)\ Площадь\ основания:\]
\[\ S_{осн} = \pi r^{2} = \frac{45}{7}\text{π\ }дм^{2}\text{.\ }\]
\[5)\ По\ теореме\ Пифагора:\ \]
\[h = \sqrt{l^{2} - r^{2}} = \sqrt{(6r)^{2} - r^{2}} =\]
\[= r\sqrt{35} = \sqrt{\frac{45}{7} \bullet 35} = 15\ дм.\]
\[6)\ V = \frac{1}{3}S_{осн}h = \frac{1}{3} \bullet \frac{45}{7}\pi \bullet 15 =\]
\[= \frac{225}{7}\text{π\ }дм^{3}.\]
\[Ответ:V = \frac{225}{7}\text{π\ }дм^{3}.\]