Решебник по геометрии 9 класс Атанасян ФГОС Задание 1237

 

\[\boxed{\mathbf{1237.ОК\ ГДЗ - домашка\ н}а\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\text{ABCD}A_{1}B_{1}C_{1}D_{1} - куб;\]

\[\textbf{а)}\ AC = 12\ см;\]

\[\textbf{б)}\ AC_{1} = 3\sqrt{2};\]

\[\textbf{в)}\ DE = 1\ см;\]

\[E \in AB;\]

\[AE = EB.\]

\[\mathbf{Найти:}\]

\[V - ?\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ 1)\ Диагональ\ грани:\ \]

\[AC = \sqrt{a^{2} + a^{2}} = a\sqrt{2}\]

\[a = \frac{\text{AC}}{\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2}\ (см).\]

\[2)\ V = a^{3} = \left( 6\sqrt{2} \right)^{3} =\]

\[= 432\sqrt{2}\ \left( см^{3} \right).\]

\[\textbf{б)}\ 1)\ Диагональ\ куба:\ \]

\[AC_{1} = \sqrt{a^{2} + a^{2} + a^{2}} = a\sqrt{3}\]

\[a = \frac{\text{AC}}{\sqrt{3}} = \frac{3\sqrt{2}}{\sqrt{3}} = \sqrt{6}.\]

\[2)\ V = a^{3} = \left( \sqrt{6} \right)^{3} = 6\sqrt{6}.\]

\[\textbf{в)}\ 1)\ DE = \sqrt{a^{2} + \left( \frac{a}{2} \right)^{2}} =\]

\[= \frac{\sqrt{5a^{2}}}{2} = \frac{a\sqrt{5}}{2} \Longrightarrow a = \frac{2 \cdot DE}{\sqrt{5}} =\]

\[= \frac{2 \cdot 1}{\sqrt{5}} = \frac{2}{\sqrt{5}}\ (см).\]

\[2)\ V = a^{3} = \left( \frac{2}{\sqrt{5}} \right)^{3} = \frac{8}{5\sqrt{5}} =\]

\[= \frac{8\sqrt{5}}{25}\ (см)^{3}.\]

\[Ответ:\ а)\ 432\sqrt{2}\ см^{3};б)\ 6\sqrt{6};\ \]

\[\textbf{в)}\frac{8\sqrt{5}}{25}\ см^{3}.\]