\[\boxed{\mathbf{841.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[K = CK \cap AB;\ \]
\[M = CK \cap AD;\]
\[S_{\text{KBC}} = S_{1};\ \ \]
\[S_{\text{CDM}} = S_{2}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}AMK\sim\mathrm{\Delta}DMC - по\ двум\ \]
\[углам:\]
\[\angle AMK =\]
\[= \angle DMC\ (как\ вертикальные\ углы);\]
\[AB \parallel CD;\ \ AD - секущая \Longrightarrow\]
\[\Longrightarrow \angle KAM = \angle CDM - как\ \]
\[накрест\ лежащие.\]
\[2)\ k = \frac{\text{AM}}{\text{MD}} = \frac{\text{AK}}{\text{DC}};\ \frac{S_{\text{AMK}}}{S_{\text{DCM}}} = k^{2};\ \ \ \]
\[S_{\text{AMK}} = k^{2} \bullet S_{2}.\]
\[3)\ \ \mathrm{\Delta}AMK\sim\mathrm{\Delta}KBC;\]
\[так\ как\ AM \parallel BC.\ \]
\[\frac{\text{KA}}{\text{KB}} = \frac{\text{KA}}{KA + AB} = \frac{1}{1 + \frac{\text{AB}}{\text{KA}}} =\]
\[= \frac{1}{1 + \frac{1}{k}} = \frac{k}{k + 1}.\]
\[S_{\text{AMK}} = \left( \frac{k}{k + 1} \right)^{2} \bullet S_{1}.\]
\[4)\ Запишем\ уравнение:\]
\[S_{\text{AMK}} = k^{2} \bullet S_{2} = \left( \frac{k}{k + 1} \right)^{2} \bullet S_{1}\]
\[(k + 1)^{2} = \frac{S_{1}}{S_{2}}\]
\[k + 1 = \sqrt{\frac{S_{1}}{S_{2}}}\]
\[k = \sqrt{\frac{S_{1}}{S_{2\ }}} - 1.\]
\[5)\ Найдем\ площадь\ \]
\[параллелограмма:\]
\[S_{\text{ABCD}} = S_{1} - S_{\text{AMK}} + S_{2} =\]
\[= S_{1} + S_{2} - \left( \sqrt{\frac{S_{1}}{S_{2\ }}} - 1 \right)^{2} \bullet S_{2} =\]
\[= S_{1} + S_{2} - \left( \frac{S_{1}}{S_{2}} - 2\sqrt{\frac{S_{1}}{S_{2\ }}} + 1 \right) \bullet S_{2} =\]
\[= S_{1} + S_{2} - S_{1} + 2\sqrt{S_{1}S_{2}} - S_{2} =\]
\[= 2\sqrt{S_{1}S_{2}}.\]
\[Ответ:S_{\text{ABCD}} = 2\sqrt{S_{1}S_{2}}.\]