\[\boxed{\mathbf{492.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC};\]
\[\text{AB} = \text{BC} = 10\ см;\]
\[\text{AC} = 12\ см.\]
\[\mathbf{Найти:}\]
\[\text{AM};\text{CL};\text{BN}.\]
\[\mathbf{Решение.}\]
\[1)\ По\ формуле\ Герона\ \]
\[\left( p = \frac{a + b + c}{2} \right):\]
\[S_{\text{ABC}} = \sqrt{p(p - a)(p - b)(p - c)};\]
\[p = \frac{10 + 10 + 12}{2} = 16\ см.\]
\[S_{\text{ABC}} =\]
\[= \sqrt{16(16 - 10)(16 - 10)(16 - 12)} =\]
\[= 48\ см^{2}.\]
\[2)\ S_{\text{ABC}} = \frac{1}{2} \bullet \text{BC} \bullet \text{AM} = 48\]
\[\frac{1}{2} \bullet 10 \bullet \text{AM} = 48\]
\[\text{AM} = \frac{48}{5} = 9\frac{3}{5} = 9,6\ см.\]
\[3)\ S_{\text{ABC}} = \frac{1}{2} \bullet \text{AB} \bullet \text{LC} = 48\]
\[\frac{1}{2} \bullet 10 \bullet \text{LC} = 48\]
\[\text{LC} = \frac{48}{5} = 9\frac{3}{5} = 9,6\ см.\]
\[4)\ S_{\text{ABC}} = \frac{1}{2} \bullet \text{AC} \bullet \text{BN} = 48\]
\[\frac{1}{2} \bullet 12 \bullet \text{BN} = 48\]
\[\text{BN} = \frac{48}{6} = 8\ см.\]
\[\mathbf{Ответ}:\text{AM} = 9\ см;\text{LC} = 9\ см;\]
\[\text{BN} = 8\ см\mathbf{.}\]