\[\boxed{\mathbf{515.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC} - равнобедренный;\]
\[\text{AB} = \text{BC};\]
\[\textbf{а)}\ \text{AB} = 20\ см;\]
\[\angle A = 30{^\circ};\]
\[\textbf{б)}\ \text{AH} = 6\ см;\]
\[\angle\text{HAC} = 45{^\circ}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}} - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ ⊿\text{BMA} - прямоугольный\ \]
\[\left( так\ как\ \text{BM}\bot\text{AC} \right):\]
\[\text{AC} = \text{AM} + \text{MC};\]
\[\text{AC} = 20\sqrt{3}\ см.\]
\[3)\ S_{\text{ABC}} = \frac{1}{2}\text{AC} \bullet \text{BM} =\]
\[= \frac{1}{2} \bullet 20\sqrt{3} \bullet 10 = 100\sqrt{3}\ см^{2}.\]
\[\textbf{б)}\ 1)\ ⊿\text{AHC} - прямоугольный.\ \]
\[По\ свойству\ прямоугольного\ \]
\[треугольника:\]
\[\angle C = 90{^\circ} - \angle\text{HAC} =\]
\[= 90{^\circ} - 45{^\circ} = 45{^\circ}.\]
\[2)\ \angle C = \angle\text{HAC} = 45{^\circ} \Longrightarrow\]
\[\Longrightarrow ⊿\text{AHC} - равнобедренный;\]
\[3)\ AC^{2} = AH^{2} + HC^{2}\]
\[AC^{2} = 36 + 36 = 72\]
\[\text{AC} = 6\sqrt{2}\ см.\]
\[4)\ \angle C = 45{^\circ};\ \ \ \text{AB} = \text{BC}:\]
\[\text{AH} - совпадает\ с\ \text{AB} \Longrightarrow\]
\[\Longrightarrow \text{AB} = \text{AH} = 6\ см.\]
\[5)\ S_{\text{ABC}} = \frac{1}{2}\text{AH} \bullet \text{HC} = \frac{1}{2} \bullet 6 \bullet 6 =\]
\[= 18\ см^{2}.\]
\[Ответ:а)\ 100\sqrt{3}\ см^{2};\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ 18\ см^{2}.\]