Решебник по геометрии 10 класс Атанасян ФГОС 521

\[\boxed{\mathbf{521.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[AB = 3\sqrt{2}\ см;\]

\[AD = 7\ см;\]

\[\angle DAB = 45{^\circ};\]

\[\angle BDB_{1} = 45{^\circ}.\]

\[Найти:\]

\[\text{V.}\]

\[Решение.\]

\[1)\ По\ теореме\ косинусов\ \]

\[(в\ \mathrm{\Delta}ABD):\]

\[BD^{2} =\]

\[= AD^{2} + AB^{2} - 2AD \bullet AB \bullet \cos{\angle A};\]

\[BD^{2} =\]

\[= 49 + 18 - 2 \bullet 7 \bullet 3\sqrt{2} \bullet \frac{1}{\sqrt{2}} =\]

\[= 49 + 18 - 42 = 25;\]

\[BD = \sqrt{25} = 5\ см.\]

\[2)\ \mathrm{\Delta}BB_{1}D - прямоугольный:\]

\[\angle B_{1} = 90{^\circ} - 45{^\circ} = 45{^\circ};\]

\[BB_{1} = BD = 5\ см.\]

\[3)\ S_{осн} = AD \bullet AB \bullet \sin{\angle BAD} =\]

\[= 7 \bullet 3\sqrt{2} \bullet \frac{1}{\sqrt{2}} = 21\ см^{2}.\]

\[4)\ V = S_{осн} \bullet BB_{1} = 21 \bullet 5 =\]

\[= 105\ см^{3}.\]

\[Ответ:\ \ V = 105\ см^{3}.\]