Решебник по геометрии 10 класс Атанасян ФГОС 522

\[\boxed{\mathbf{522.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[BD_{1}\bot A_{1}C;\]

\[BD_{1} = 6\ см;\]

\[A_{1}C = 8\ см;\]

\[AB = 3\ см.\]

\[Найти:\]

\[\text{V.}\]

\[Решение.\]

\[1)\ A_{1}D_{1}CB - ромб:\]

\[BD_{1}\bot A_{1}C;\ \ \ \]

\[A_{1}B = CD_{1};\ \]

\[A_{1}D_{1} = BC.\]

\[Отсюда:\ \]

\[BO = OD_{1} = 3\ см;\]

\[A_{1}O = OC = 4\ см.\]

\[2)\ \mathrm{\Delta}A_{1}OB - прямоугольный:\]

\[A_{1}B = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5\ см.\]

\[3)\ \mathrm{\Delta}A_{1}AB - прямоугольный:\]

\[AA_{1} = \sqrt{25 - 3^{2}} = \sqrt{16} = 4\ см.\]

\[4)\ \mathrm{\Delta}A_{1}AC - прямоугольный:\]

\[AC = \sqrt{8^{2} - 16} = \sqrt{48} = 4\sqrt{3}\ см.\]

\[5)\ По\ теореме\ косинусов\ \]

\[(в\ \mathrm{\Delta}ABC):\]

\[\left( 4\sqrt{3} \right)^{2} =\]

\[= 3^{2} + 5^{2} - 2 \bullet 3 \bullet 5\cos{\angle B};\]

\[\cos{\angle B} = - \frac{7}{15}.\]

\[\sin{\angle B} = \sqrt{1 - \cos^{2}{\angle B}} =\]

\[= \sqrt{1 - \frac{49}{225}} = \frac{\sqrt{176}}{15}.\]

\[6)\ S_{осн} = AB \bullet BC \bullet \sin{\angle B} =\]

\[= 3 \bullet 5 \bullet \frac{\sqrt{176}}{15} = \sqrt{176}\ см^{2}.\]

\[7)\ V = S_{осн} \bullet AA_{1} = \sqrt{176} \bullet 4 =\]

\[= 4 \bullet \sqrt{16 \bullet 11} = 16\sqrt{11}\ см^{3}.\]

\[Ответ:\ \ V = 6\sqrt{11}\ см^{3}.\]