ГДЗ по геометрии 8 класс Атанасян Задание 841

\[\boxed{\mathbf{841.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - параллелограмм;\]

\[K = CK \cap AB;\ \]

\[M = CK \cap AD;\]

\[S_{\text{KBC}} = S_{1};\ \ \]

\[S_{\text{CDM}} = S_{2}.\]

\[\mathbf{Найти:}\]

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

\[\mathbf{Решение.}\]

\[1)\ \mathrm{\Delta}AMK\sim\mathrm{\Delta}DMC - по\ двум\ \]

\[углам:\]

\[\angle AMK =\]

\[= \angle DMC\ (как\ вертикальные\ углы);\]

\[AB \parallel CD;\ \ AD - секущая \Longrightarrow\]

\[\Longrightarrow \angle KAM = \angle CDM - как\ \]

\[накрест\ лежащие.\]

\[2)\ k = \frac{\text{AM}}{\text{MD}} = \frac{\text{AK}}{\text{DC}};\ \frac{S_{\text{AMK}}}{S_{\text{DCM}}} = k^{2};\ \ \ \]

\[S_{\text{AMK}} = k^{2} \bullet S_{2}.\]

\[3)\ \ \mathrm{\Delta}AMK\sim\mathrm{\Delta}KBC;\]

\[так\ как\ AM \parallel BC.\ \]

\[\frac{\text{KA}}{\text{KB}} = \frac{\text{KA}}{KA + AB} = \frac{1}{1 + \frac{\text{AB}}{\text{KA}}} =\]

\[= \frac{1}{1 + \frac{1}{k}} = \frac{k}{k + 1}.\]

\[S_{\text{AMK}} = \left( \frac{k}{k + 1} \right)^{2} \bullet S_{1}.\]

\[4)\ Запишем\ уравнение:\]

\[S_{\text{AMK}} = k^{2} \bullet S_{2} = \left( \frac{k}{k + 1} \right)^{2} \bullet S_{1}\]

\[(k + 1)^{2} = \frac{S_{1}}{S_{2}}\]

\[k + 1 = \sqrt{\frac{S_{1}}{S_{2}}}\]

\[k = \sqrt{\frac{S_{1}}{S_{2\ }}} - 1.\]

\[5)\ Найдем\ площадь\ \]

\[параллелограмма:\]

\[S_{\text{ABCD}} = S_{1} - S_{\text{AMK}} + S_{2} =\]

\[= S_{1} + S_{2} - \left( \sqrt{\frac{S_{1}}{S_{2\ }}} - 1 \right)^{2} \bullet S_{2} =\]

\[= S_{1} + S_{2} - \left( \frac{S_{1}}{S_{2}} - 2\sqrt{\frac{S_{1}}{S_{2\ }}} + 1 \right) \bullet S_{2} =\]

\[= S_{1} + S_{2} - S_{1} + 2\sqrt{S_{1}S_{2}} - S_{2} =\]

\[= 2\sqrt{S_{1}S_{2}}.\]

\[Ответ:S_{\text{ABCD}} = 2\sqrt{S_{1}S_{2}}.\]