ГДЗ по геометрии 7 класс Атанасян ФГОС Задание 631

\[\boxed{\mathbf{631.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - ромб;\]

\[BH = \frac{4\sqrt{2}}{6}\ см;\]

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

\[\mathbf{Найти:}\]

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

\[\mathbf{Решение.}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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