ГДЗ по геометрии 11 класс. Атанасян ФГОС 335

\[\boxed{\mathbf{335.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\angle ACB = 90{^\circ};\]

\[h - высота\ и\ образующая;\]

\[R - радиус\ цилиндра.\]

\[Найти:\]

\[S_{сеч}.\]

\[Решение.\]

\[По\ теореме\ Пифагора:\]

\[AC = \sqrt{(2R)^{2} - BC^{2}} =\]

\[= \sqrt{4R^{2} - BC^{2}}.\]

\[S_{1} = S_{BB_{1}C_{1}C} = BC \cdot h = S_{2} =\]

\[= S_{\text{AC}C_{1}A_{1}} = h \cdot \sqrt{4R^{2} - BC^{2}}.\]

\[BC \cdot h = h\sqrt{4R^{2} - BC^{2}}\]

\[BC = \sqrt{4R^{2} - BC^{2}}\]

\[BC^{2} = 4R^{2} - BC^{2}\]

\[2BC^{2} = 4R^{2}\]

\[BC^{2} = 2R^{2}\]

\[BC = R\sqrt{2}\]

\[R = \frac{\text{BC}}{\sqrt{2}}.\]

\[S = 2Rh = \frac{2h \cdot BC}{\sqrt{2}} = \sqrt{2}BC \cdot h =\]

\[= \sqrt{2}S_{1} = \sqrt{2}S_{2}.\]

\[Ответ:S\sqrt{2}.\]