ГДЗ по геометрии 9 класс Атанасян Задание 526

\[\boxed{\mathbf{526.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\text{ABCD} - ромб;\]

\[\text{BH} = \frac{4\sqrt{2}}{6}\ см;\]

\[\frac{2}{3}\text{AC} = \frac{4\sqrt{2}}{6}\ см.\]

\[\mathbf{Найти:}\]

\[S_{\text{ABCD}} - ?\]

\[\mathbf{Решение.}\]

\[1)\ \frac{2}{3}\text{AC} = \frac{4\sqrt{2}}{6}\]

\[\text{AC} = \frac{4\sqrt{2}}{6} \bullet \frac{3}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}.\]

\[2)\ S_{\text{ABCD}} = \frac{1}{2}\text{BD} \bullet \text{AC} =\]

\[= \text{BH} \bullet \text{AD} = \frac{2}{3}\text{AC} \bullet \text{AD};\]

\[\frac{1}{2}\text{BD} = \frac{2}{3}\text{AD}\]

\[\text{BD} = \frac{4}{3}\text{AD}\]

\[\frac{\text{BD}}{2} = \frac{2}{3}\text{AD}.\]

\[3)\ \mathrm{\Delta}\text{AOD} - прямоугольный:\]

\[AD^{2} = AO^{2} + OD^{2} =\]

\[= \left( \frac{\sqrt{2}}{2} \right)^{2} + \left( \frac{2}{3}\text{AD} \right)^{2}\]

\[AD^{2} = \frac{2}{4} + \frac{4}{9}AD^{2}\]

\[AD^{2} - \frac{4}{9}AD^{2} = \frac{1}{2}\]

\[\frac{5}{9}AD^{2} = \frac{1}{2}\]

\[AD^{2} = \frac{9}{10}\]

\[\text{AD} = \frac{3}{\sqrt{10}}.\]

\[4)\ S_{\text{ABCD}} = \frac{4\sqrt{2}}{6} \bullet \frac{3}{\sqrt{10}} = \frac{2\sqrt{2}}{\sqrt{10}} =\]

\[= \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\ см^{2}.\]

\[\mathbf{Ответ:}\frac{2\sqrt{5}}{5}\ см^{2}.\]