\[\boxed{\mathbf{630.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[M \in ABC;\]
\[MH = 6\ см;\]
\[ME = 2\ см;\]
\[AB = 13\ см;\]
\[BC = 14\ см;\]
\[AC = 15\ см.\]
\[\mathbf{Найти:}\]
\[MF - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABC}} = S_{\text{ABM}} + S_{\text{BMC}} + S_{\text{AMC}}.\]
\[2)\ По\ формуле\ Герона:\]
\[p = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} =\]
\[= \frac{42}{2} = 21\ см.\]
\[= \sqrt{21 \bullet 8 \bullet 7 \bullet 6} =\]
\[= \sqrt{7 \bullet 3 \bullet 4 \bullet 2 \bullet 7 \bullet 2 \bullet 3} =\]
\[= 7 \bullet 3 \bullet 2 \bullet 2 = 84\ см^{2}.\]
\[4)\ S_{\text{ABM}} = \frac{1}{2}MH \bullet AB =\]
\[= \frac{1}{2} \bullet 6 \bullet 13 = 39\ см^{2}.\]
\[S_{\text{AMC}} = \frac{1}{2}ME \bullet AC = \frac{1}{2} \bullet 2 \bullet 15 =\]
\[= 15\ см^{2}.\]
\[5)\ S_{\text{BMC}} = S_{\text{ABC}} - S_{\text{ABM}} - S_{\text{AMC}} =\]
\[= 84 - 39 - 15 = 30\ см^{2}.\]
\[6)\ S_{\text{BMC}} = \frac{1}{2}MF \bullet BC =\]
\[= \frac{1}{2} \bullet 14 \bullet MF = 30\ см^{2}\]
\[7MF = 30\]
\[MF = \frac{30}{7} = 4\frac{2}{7}\ см.\]
\[\mathbf{Ответ:\ }4\frac{2}{7}\ см.\]