Решебник по геометрии 10 класс Атанасян ФГОС 231

\[\boxed{\mathbf{231.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[A_{1}B_{1} = 15\ см;\]

\[A_{1}D_{1} = 8\ см;\]

\[\angle B_{1}A_{1}D_{1} = 60{^\circ};\]

\[S_{BB_{1}D_{1}D} = 130\ см^{2}.\]

\[Найти:\]

\[S_{пов}.\]

\[Решение.\]

\[По\ теореме\ косинусов:\]

\[BD^{2} =\]

\[= AB^{2} + AD^{2} - 2AB \cdot AD \cdot \cos{60{^\circ}} =\]

\[= 64 + 225 - 2 \cdot 8 \cdot 15 \cdot \frac{1}{2} = 169;\]

\[BD = 13\ см.\]

\[AC^{2} =\]

\[= AB^{2} + BC^{2} - 2AB \cdot BC \cdot \cos{120{^\circ}} =\]

\[= 64 + 225 + 2 \cdot 8 \cdot 15 \cdot \frac{1}{2} = 409;\]

\[AC = \sqrt{409}.\]

\[AC > BD:\]

\[BD - меньшая\ диагональ;\]

\[BB_{1}D_{1}D - меньшее\ сечение.\]

\[S_{BB_{1}D_{1}D} = BB_{1} \cdot BD =\]

\[= BB_{1} \cdot 13 = 130\]

\[BB_{1} = 10\ см.\]

\[S_{пов} = 2S_{осн} + S_{бок};\ \ \ \]

\[S_{бок} = 2S_{AA_{1}D_{1}D} + 2S_{AA_{1}B_{1}B} =\]

\[= 2AD \cdot AA_{1} + 2AB \cdot AA_{1} =\]

\[= 2 \cdot 15 \cdot 10 + 2 \cdot 8 \cdot 10 =\]

\[= 460\ см^{2}.\]

\[S_{осн} = AB \cdot AD \cdot \sin{\angle A} =\]

\[= 8 \cdot 15 \cdot \sin{60{^\circ}} = 60\sqrt{3}\ см^{2}.\]

\[S_{пов} = 2 \cdot 60\sqrt{3} + 460 =\]

\[= 20 \cdot \left( 23 + 6\sqrt{3} \right)\ см^{2}.\]

\[Ответ:\ 20 \cdot \left( 23 + 6\sqrt{3} \right)\ см^{2}.\]