ГДЗ по геометрии 9 класс Атанасян Задание 1193

\[\boxed{\mathbf{1193.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

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

\[параллелепипед;\]

\[\textbf{а)}\ AB = BC = 1;\]

\[CC_{1} = 2;\]

\[\textbf{б)}\ AB = 8;\]

\[BC = 9;\]

\[CC_{1} = 12;\]

\[\textbf{в)}\ AB = \sqrt{39};\]

\[BC = 7;\]

\[CC_{1} = 9.\]

\[\mathbf{Найти:}\]

\[AC_{1} - ?\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ 1)\ По\ теореме\ Пифагора:\]

\[AC = \sqrt{AB^{2} + BC^{2}} =\]

\[= \sqrt{1^{2} + 1^{2}} = \sqrt{2}.\]

\[2)\ AC_{1} = \sqrt{AC^{2} + \left( CC_{1} \right)^{2}} =\]

\[= \sqrt{2 + 2^{2}} = \sqrt{6}.\]

\[\textbf{б)}\ 1)\ По\ теореме\ Пифагора:\]

\[AC = \sqrt{AB^{2} + BC^{2}} =\]

\[= \sqrt{8^{2} + 9^{2}} = \sqrt{145}.\]

\[2)\ AC_{1} = \sqrt{AC^{2} + \left( CC_{1} \right)^{2}} =\]

\[= \sqrt{145 + 12^{2}} = \sqrt{289} = 17.\]

\[\textbf{в)}\ 1)\ По\ теореме\ Пифагора:\]

\[AC = \sqrt{AB^{2} + BC^{2}} =\]

\[= \sqrt{39 + 7^{2}} = \sqrt{88}.\]

\[2)\ AC_{1} = \sqrt{AC^{2} + \left( CC_{1} \right)^{2}} =\]

\[= \sqrt{88 + 81} = \sqrt{169} = 13.\]

\[Ответ:а)\ \sqrt{6};\ б)\ 17;\ в)\ 13.\]