ГДЗ по геометрии 9 класс Атанасян Задание 438

\[\boxed{\mathbf{438.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\text{ABCD} - трапеция;\]

\[\text{AC}\bot\text{CD};\]

\[\angle\text{BAC} = \angle\text{CAD};\]

\[\angle D = 60{^\circ};\]

\[P_{\text{ABCD}} = 20\ см.\]

\[\mathbf{Найти:}\]

\[\text{AD} - ?\]

\[\mathbf{Решение.}\]

\[1)\ \angle\text{CAD} = 90{^\circ} - 60{^\circ} = 30{^\circ}\]

\[По\ свойству\ прямоугольного\ \]

\[треугольника:\]

\[\text{CD} = \frac{1}{2}\text{AD}\text{.\ }\]

\[2)\ \angle\text{BAC} = \angle\text{CAD} = 30{^\circ}.\]

\[3)\ \angle A = \angle\text{BAC} + \angle\text{CAD} =\]

\[= 30 + 30 = 60{^\circ}\]

\[\angle A = \angle D = 60{^\circ}.\]

\[Следовательно:\]

\[\text{ABCD} - равнобедренная\ \]

\[трапеци \Longrightarrow \text{AB} = \text{CD}.\]

\[4)\ \text{ABCD} - трапеция \Longrightarrow\]

\[\Longrightarrow \text{BC} \parallel \text{AD}.\]

\[5)\ \text{BC} \parallel \text{AD}\ и\ \text{AC} - секущая:\ \]

\[\angle\text{BCA} = \angle\text{CAD} =\]

\[= 30{^\circ}\ (как\ накрестлежащие).\]

\[6)\ \angle\text{BAC} = \angle\text{BCA} = 30{^\circ}:\]

\[\mathrm{\Delta}\text{ABC} - равнобедренный \Longrightarrow\]

\[\Longrightarrow \ \text{AB} = \text{BC}.\]

\[7)\ P_{\text{ABCD}} =\]

\[= \text{AB} + \text{BC} + \text{CD} + \text{AD} = 20\ см;\]

\[\text{AB} = \text{BC} = \text{CD};\]

\[P_{\text{ABCD}} = 3\text{AB} + \text{AD} = 20\ см;\]

\[\text{CD} = \frac{1}{2}\text{AD};\]

\[P_{\text{ABCD}} = 3 \bullet \frac{1}{2}\text{AD} + \text{AD} = 20\ см;\]

\[\frac{5}{2}\text{AD} = 20\ см\]

\[\text{AD} = 8\ см.\]

\[Ответ:\text{AD} = 8\ см.\]