ГДЗ по геометрии 8 класс Атанасян Задание 834

\[\boxed{\mathbf{834.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - трапеция;\]

\[\text{AD} \parallel BC;\ \ \]

\[O = AC \cap BD;\]

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

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

\[\mathbf{Найти:}\]

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

\[\mathbf{Решение.}\]

\[1)\ AB \parallel CD;\ \ BD - секущая:\ \]

\[\ \angle CBO =\]

\[= \angle ADO\ (накрест\ лежащие).\]

\[\mathrm{\Delta}BOC\sim\mathrm{\Delta}DOA - по\ двум\ углам:\]

\[\angle BOC =\]

\[= \angle DOA\ (как\ вертикальные).\]

\[2)\ Коэффициент\ подобия\ \]

\[треугольников:\]

\[k^{2} = \frac{S_{\text{BOC}}}{S_{\text{AOD}}} = \frac{S_{1}}{S_{2}} \Longrightarrow k = \sqrt{\frac{S_{1}}{S_{2}}}.\]

\[\frac{S_{\text{DAO}}}{S_{\text{DCO}}} = \frac{\text{AO}}{\text{CO}} = \frac{1}{k}\]

\[S_{\text{DAC}} = S_{\text{DAO}} + S_{\text{DCO}} =\]

\[= S_{\text{DAO}} + k \bullet S_{\text{DAO}} = (1 + k) \bullet S_{2}.\]

\[\frac{S_{\text{BAO}}}{S_{\text{BCO}}} = \frac{\text{AO}}{\text{CO}} = \frac{1}{k}\]

\[S_{\text{BAC}} = S_{\text{BAO}} + S_{\text{BCO}} =\]

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

\[3)\ Получаем:\]

\[S_{\text{ABCD}} = S_{\text{DAC}} + S_{\text{BAC}} =\]

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

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

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

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

\[Ответ:S_{\text{ABCD}} = \left( \sqrt{S_{1}} + \sqrt{S_{2}} \right)^{2}.\]