Решебник по геометрии 11 класс. Атанасян ФГОС 242

\[\boxed{\mathbf{242.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[ABCD - квадрат.\]

\[Найти:\]

\[\textbf{а)}\ h;\]

\[\textbf{б)}\ S_{бок}.\]

\[Решение.\]

\[1)\ AC - диагональ;\ \ AC > AD:\]

\[проекция\ AC\ наклонной\ \text{SC\ }\]

\[больше\ проекций\ \text{AD\ }и\ \text{AB\ }\]

\[наклонных\ SD\ и\ \text{SB.}\]

\[2)\ SC = 12\ см;\ \ AC = AB\sqrt{2};\ \ \]

\[SA = AB.\]

\[3)\ По\ теореме\ Пифагора:\]

\[SC = \sqrt{SA^{2} + AC^{2}} =\]

\[= \sqrt{AB^{2} + 2AB^{2}} = AB\sqrt{3}.\]

\[\text{AB}\sqrt{3} = 12\]

\[AB = \frac{12\sqrt{3}}{3} = 4\sqrt{3}\ см.\]

\[\textbf{а)}\ SA = 4\sqrt{3}\ см.\]

\[\textbf{б)}\ S_{\text{SAD}} = S_{\text{SAB}} = \frac{1}{2}SA \cdot AB =\]

\[= \frac{1}{2}AB^{2} = \frac{1}{2} \cdot \left( 4\sqrt{3} \right)^{2} = 24\ см^{2}.\]

\[По\ теореме\ о\ трех\ \]

\[перпендикулярах:\]

\[SD\bot DC;\ \ SB\bot BC.\]

\[По\ теореме\ Пифагора:\]

\[SD = \sqrt{2AB^{2}} = \sqrt{2 \cdot 16 \cdot 3} =\]

\[= 4\sqrt{6}\ см.\]

\[S_{\text{SDC}} = \frac{1}{2}SD \cdot DC = \frac{48\sqrt{2}\ }{2} =\]

\[= 24\sqrt{2}\ см^{2};\]

\[S_{\text{SBC}} = \frac{1}{2}SB \cdot BC = \frac{48\sqrt{2}}{2} =\]

\[= 24\sqrt{2}\ см^{2};\]

\[S_{бок} = 2 \cdot 24 + 2\sqrt{2} \cdot 24 =\]

\[= 48 \cdot \left( \sqrt{2} + 1 \right)\ см^{2}.\]

\[Ответ:4\sqrt{3}\ см;\ \ \]

\[48 \cdot \left( \sqrt{2} + 1 \right)\ см^{2}.\]