\[\boxed{\mathbf{525.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC};\]
\[M \in \text{ABC};\]
\[\text{MH} = 6\ см;\]
\[\text{ME} = 2\ см;\]
\[\text{AB} = 13\ см;\]
\[\text{BC} = 14\ см;\]
\[\text{AC} = 15\ см.\]
\[\mathbf{Найти:}\]
\[\text{MF} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABC}} = S_{\text{ABM}} + S_{\text{BMC}} + S_{\text{AMC}}.\]
\[2)\ По\ формуле\ Герона:\]
\[p = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} =\]
\[= \frac{42}{2} = 21\ см.\]
\[3)\ S_{\text{ABC}} =\]
\[= \sqrt{21(21 - 13)(21 - 14)(21 - 15)} =\]
\[= \sqrt{21 \bullet 8 \bullet 7 \bullet 6} =\]
\[= \sqrt{7 \bullet 3 \bullet 4 \bullet 2 \bullet 7 \bullet 2 \bullet 3} =\]
\[= 7 \bullet 3 \bullet 2 \bullet 2 = 84\ см^{2}.\]
\[4)\ S_{\text{ABM}} = \frac{1}{2}\text{MH} \bullet \text{AB} =\]
\[= \frac{1}{2} \bullet 6 \bullet 13 = 39\ см^{2}.\]
\[S_{\text{AMC}} = \frac{1}{2}\text{ME} \bullet \text{AC} = \frac{1}{2} \bullet 2 \bullet 15 =\]
\[= 15\ см^{2}.\]
\[5)\ S_{\text{BMC}} = S_{\text{ABC}} - S_{\text{ABM}} - S_{\text{AMC}} =\]
\[= 84 - 39 - 15 = 30\ см^{2}.\]
\[6)\ S_{\text{BMC}} = \frac{1}{2}\text{MF} \bullet \text{BC} =\]
\[= \frac{1}{2} \bullet 14 \bullet \text{MF} = 30\ см^{2}\]
\[7\text{MF} = 30\]
\[\text{MF} = \frac{30}{7} = 4\frac{2}{7}\ см.\]
\[\mathbf{Ответ:\ }4\frac{2}{7}\ см.\]