\[\boxed{\mathbf{834.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - трапеция;\]
\[\text{AD} \parallel BC;\ \ \]
\[O = AC \cap BD;\]
\[S_{\text{BOC}} = S_{1};\]
\[S_{\text{AOD}} = S_{2}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ AB \parallel CD;\ \ BD - секущая:\ \]
\[\ \angle CBO =\]
\[= \angle ADO\ (накрест\ лежащие).\]
\[\mathrm{\Delta}BOC\sim\mathrm{\Delta}DOA - по\ двум\ углам:\]
\[\angle BOC =\]
\[= \angle DOA\ (как\ вертикальные).\]
\[2)\ Коэффициент\ подобия\ \]
\[треугольников:\]
\[k^{2} = \frac{S_{\text{BOC}}}{S_{\text{AOD}}} = \frac{S_{1}}{S_{2}} \Longrightarrow k = \sqrt{\frac{S_{1}}{S_{2}}}.\]
\[\frac{S_{\text{DAO}}}{S_{\text{DCO}}} = \frac{\text{AO}}{\text{CO}} = \frac{1}{k}\]
\[S_{\text{DAC}} = S_{\text{DAO}} + S_{\text{DCO}} =\]
\[= S_{\text{DAO}} + k \bullet S_{\text{DAO}} = (1 + k) \bullet S_{2}.\]
\[\frac{S_{\text{BAO}}}{S_{\text{BCO}}} = \frac{\text{AO}}{\text{CO}} = \frac{1}{k}\]
\[S_{\text{BAC}} = S_{\text{BAO}} + S_{\text{BCO}} =\]
\[= \frac{S_{\text{BCO}}}{k} + S_{\text{BCO}} = \left( \frac{1}{k} + 1 \right) \bullet S_{1}.\]
\[3)\ Получаем:\]
\[S_{\text{ABCD}} = S_{\text{DAC}} + S_{\text{BAC}} =\]
\[= (1 + k) \bullet S_{2} + \left( \frac{1}{k} + 1 \right) \bullet S_{1} =\]
\[= S_{2} + \sqrt{\frac{S_{1}}{S_{2}}} \bullet S_{2} + \sqrt{\frac{S_{2}}{S_{1}}} \bullet S_{1} + S_{1} =\]
\[= S_{1} + 2\sqrt{S_{1}S_{2}} + S_{2} =\]
\[= \left( \sqrt{S_{1}} + \sqrt{S_{2}} \right)^{2}.\]
\[Ответ:S_{\text{ABCD}} = \left( \sqrt{S_{1}} + \sqrt{S_{2}} \right)^{2}.\]