\[\boxed{\mathbf{290.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[параллелепипед;\]
\[AC = l;\]
\[\angle CAD = \varphi;\ \ \]
\[\angle C_{1}AD = \theta.\]
\[Найти:\]
\[S_{бок}.\]
\[Решение.\ \]
\[AD = AC \cdot \cos{\angle CAD} = l \cdot \cos\varphi;\]
\[CD = l \cdot \sin\varphi;\]
\[AC_{1} = \frac{\text{AD}}{\cos{\angle C_{1}\text{AD}}} = \frac{l \cdot \cos\varphi}{\cos\theta};\]
\[CC_{1} = DD_{1} = \sqrt{AC^{2} - AC_{1}^{2}} =\]
\[= l \cdot \sqrt{\frac{\text{co}s^{2}\varphi}{\text{co}s^{2}\theta} - 1}.\]
\[S_{бок} =\]
\[= 2 \cdot \left( AD \cdot DD_{1} + CD \cdot DD_{1} \right) =\]