Решебник по геометрии 10 класс Атанасян ФГОС 264

\[\boxed{\mathbf{264.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[AB = a;\]

\[S_{\text{AMB}} = S_{\text{AMD}}.\]

\[Найти:\ \]

\[S_{бок}.\]

\[Решение.\]

\[AB = a;\]

\[AD = 2a.\]

\[Пирамида\ правильная:\]

\[S_{бок} = 6S_{\text{AMB}}.\]

\[S_{\text{AMB}} = \frac{1}{2}AB \cdot MK;\]

\[S_{\text{AMD}} = \frac{1}{2}MO \cdot AD:\]

\[\frac{1}{2}AB \cdot MK = \frac{1}{2}MO \cdot AD\]

\[\frac{1}{2}MO \cdot 2a = \frac{1}{2}a \cdot MK\]

\[MO = \frac{1}{2}\text{MK.}\]

\[⊿MOK - прямоугольный:\]

\[\angle MKO = 30{^\circ};\]

\[\cos{\angle MKO} = \frac{\text{KO}}{\text{MK}}\]

\[MK = \frac{\text{KO}}{\cos{30{^\circ}}}.\]

\[В\ треугольнике\ AKO:\]

\[AK = \frac{a}{2};\ \ AO = a;\ \ OK = \frac{a\sqrt{3}}{2};\]

\[MK = \frac{a\sqrt{3} \cdot 2}{2\sqrt{3}} = a.\]

\[S_{\text{AMB}} = \frac{1}{2}a \cdot a = \frac{a^{2}}{2};\]

\[S_{бок} = 6 \cdot \frac{a^{2}}{2} = 3a^{2}\text{.\ }\]

\[Ответ:3a^{2}.\]