\[\boxed{\mathbf{489.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC} - равносторонний;\]
\[\text{AB} = a;\]
\[\textbf{а)}\ a = 5\ см;\]
\[\textbf{б)}\ a = 1,2\ см;\]
\[\textbf{в)}\ a = 2\sqrt{2}\ дм.\]
\[\mathbf{Доказать:}\]
\[S_{\text{ABC}} = \frac{a^{2}\sqrt{3}}{4}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}\text{ABH} - прямоугольный:\]
\[AB^{2} = BH^{2} + AH^{2};\]
\[\text{AB} = a;\]
\[BH^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3a^{2}}{4} \Longleftarrow \text{BH} =\]
\[= \frac{\sqrt{3}}{2}a.\]
\[2)\ S_{\text{ABC}} = \frac{1}{2}\text{AC} \bullet \text{BH} =\]
\[= \frac{1}{2}a \bullet \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}a^{2}}{4}.\]
\[Что\ и\ требовалось\ доказать;\]
\[\textbf{а)}\ S_{\text{ABC}} = \frac{\sqrt{3} \bullet 25}{4} = \frac{25\sqrt{3}}{4}\ см^{2}.\]
\[\textbf{б)}\ S_{\text{ABC}} = \frac{1,44\sqrt{3}}{4} = 0,36\sqrt{3}\ см^{2}.\]
\[\textbf{в)}\ S_{\text{ABC}} = \frac{8\sqrt{3}}{4} = 2\sqrt{3}\ дм^{2}.\]