\[\boxed{\mathbf{516.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC};\]
\[\text{BC} = 34\ см;\]
\[\text{BM} = \text{MC};\]
\(\text{MN}\bot\text{AC};\)
\[\text{AN} = 25\ см;\]
\[\text{NC} = 15\ см.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ ⊿\text{BHC} - прямоугольный\ \]
\[\left( так\ как\ \text{BH}\bot\text{AC} \right):\]
\[\left. \ \frac{\text{BM} = \text{MC}}{\text{BH} \parallel \text{MN}} \right|\]
\[\Longrightarrow \text{HN} = \text{NC} =\]
\[= 15\ см\ (по\ теореме\ Фалесса).\]
\[2)\ \text{AC} = \text{AN} + \text{NC} = 25 + 15 =\]
\[= 40\ см.\]
\[3)\ \text{HC} = \text{HN} + \text{NC} = 15 + 15 =\]
\[= 30\ см.\]
\[4)\ BH^{2} = BC^{2} - HC^{2}\]
\[BH^{2} = 1156 - 900 = 256\]
\[\text{BH} = 16\ см.\]
\[5)\ S_{\text{ABC}} = \frac{1}{2}\text{AC} \bullet \text{BH} =\]
\[= \frac{1}{2} \bullet 40 \bullet 16 = 320\ см^{2}.\]
\[Ответ:320\ см^{2}.\]