Решебник по геометрии 10 класс Атанасян ФГОС 369

\[\boxed{\mathbf{369.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[конус;\]

\[S_{бок} = 80\ см^{2};\]

\[SO = OM;\]

\[CD\bot SM.\]

\[Найти:\ \]

\[S_{бок}.\]

\[Решение.\ \]

\[1)\ S_{бок} = \pi(r + R)l;\ \ r = OC;\ \ \]

\[R = MA;\ \ l = CA.\]

\[2)\ В\ треугольнике\ ASM:\]

\[CO - средняя\ линия.\]

\[3)\ AC = CS;\ CA = CS = l;\ \ \]

\[S_{бок} = \pi R \cdot AC:\]

\[\pi R \cdot 2l = 80\]

\[R\pi l = 40\]

\[l = \frac{40}{\text{πR}}.\]

\[4)\ ⊿SOC\ подобен\ ⊿SMA:\]

\[\frac{\text{SO}}{\text{SM}} = \frac{\text{OC}}{\text{MA}} = \frac{\text{SC}}{\text{SA}};\]

\[\frac{r}{R} = \frac{l}{2l} = \frac{1}{2}\]

\[R = 2r;\ \ \ \ r = \frac{R}{2}.\]

\[5)\ S_{бок} = \pi\left( \frac{R}{2} + R \right) \cdot \frac{40}{\text{πR}} =\]

\[= \frac{3R\pi \cdot 40}{2\pi R} = 60\ см^{2}.\]

\[Ответ:60\ см^{2}.\]