Решебник по геометрии 10 класс Атанасян ФГОС 354

\[\boxed{\mathbf{354.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[SO = h = 10\ см;\]

\[\angle COB = 60{^\circ}.\]

\[Найти:\]

\[S_{сеч}.\]

\[Решение.\]

\[BC - хорда.\]

\[{\angle SAO - двугранный\ угол\ }{между\ плоскостью\ основания\ }\]

\[и\ \text{BSC.}\]

\[По\ теореме\ о\ трех\ \]

\[перпендикулярах:\]

\[OA\bot BC;\ \ SA\bot BC;\ \ SO\bot AO.\]

\[В\ треугольнике\ COB:\]

\[BC = r = 2CA.\]

\[S_{сеч} = \frac{1}{2}BC \cdot SA.\]

\[\textbf{а)}\ \angle SAO = 30{^\circ}:\]

\[SA = \frac{h}{\sin{30{^\circ}\ }} = 2h;\]

\[\frac{\text{CA}}{\text{OA}} = tg30{^\circ} = \frac{1}{\sqrt{3}};\ \ \ CA = \frac{\text{OA}}{\sqrt{3}}.\]

\[\frac{h}{\text{OA}} = tg\ 30{^\circ} = \frac{1}{\sqrt{3}};\ \ \ OA = h\sqrt{3}.\]

\[CA = \frac{h\sqrt{3}}{\sqrt{3}} = h;\ \ \ BC = 2h:\]

\[S_{сеч} = \frac{1}{2} \cdot 2h \cdot 2h = 2h^{2} =\]

\[= 2 \cdot 10^{2} = 200\ см^{2}.\]

\[\textbf{б)}\ \angle SAO = 45{^\circ}:\]

\[SA = h\sqrt{2} = 10\sqrt{2}\ см;\]

\[OA = h;\ \ CA = \frac{h}{\sqrt{3}};\ \ BC = \frac{2h}{\sqrt{3}};\]

\[S_{сеч} = \frac{1}{2} \cdot \frac{2h}{\sqrt{3}} \cdot h\sqrt{2} = \frac{h^{2}\sqrt{2}}{\sqrt{3}} =\]

\[= \frac{100\sqrt{6}}{3}\ см^{2}.\]

\[\textbf{в)}\ \angle SAO = 60{^\circ}:\]

\[\frac{h}{\text{OA}} = tg60{^\circ} = \sqrt{3};\ \ \ OA = \frac{h}{\sqrt{3}};\]

\[CA = \frac{\text{OA}}{\sqrt{3}} = \frac{h}{3};\ \ CB = \frac{2h}{3};\ \ \]

\[SA = \frac{h}{\sin{60{^\circ}}} = \frac{2h}{\sqrt{3}}.\]

\[S_{сеч} = \frac{1}{2} \cdot \frac{2h}{\sqrt{3}} \cdot \frac{2h}{3} = \frac{2h^{2}}{3\sqrt{3}} = \frac{200}{3\sqrt{3}} =\]

\[= \frac{200\sqrt{3}}{9}\ см^{2}.\]