\[\boxed{\mathbf{528.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - трапеция;\]
\[\text{CD} = 12\ см;\]
\[\text{OF} = 5\ см;\]
\[\text{BD} \cap \text{AC} = O.\]
\[\mathbf{Найти:}\]
\[S_{\text{AOB}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{COD}} = \frac{1}{2}\text{CD} \bullet \text{OF} =\]
\[= \frac{1}{2} \bullet 12 \bullet 5 = 30\ см^{2}.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}\text{ABD}\ и\ \mathrm{\Delta}\text{ACD}:\]
\[S_{\text{ABD}} = \frac{1}{2}\text{AD} \bullet \text{BH};\]
\[S_{\text{ACD}} = \frac{1}{2}\text{AD} \bullet \text{CE};\ \]
\[\text{BH} = \text{CE}\ (как\ высоты).\]
\[Значит:\ \]
\[S_{\text{ABD}} = S_{\text{ACD}}.\]
\[3)\ S_{\text{ABD}} = S_{\text{ABO}} + S_{\text{AOD}};\]
\[S_{\text{ACD}} = S_{\text{COD}} + S_{\text{AOD}};\]
\[получаем:\]
\[S_{\text{AOB}} = S_{\text{COD}} = 30\ см^{2}.\]
\[\mathbf{Ответ:}30\ см^{2}.\]