ГДЗ по геометрии 8 класс Атанасян Задание 571

\[\boxed{\mathbf{571.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\mathbf{\ задачи:}\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}ABC;\]

\[AA_{1},\ BB_{1} - медианы;\]

\[AA_{1} \cap BB_{1} = O;\]

\[S_{\text{ABO}} = S.\]

\[\mathbf{Найти:}\]

\[S_{\text{ABC}} - ?\]

\[\mathbf{Решение.}\]

\[1)\ S_{\text{ABC}} = \frac{1}{2}AB \bullet CH;\ \ \ \]

\[S_{\text{ABO}} = \frac{1}{2}AB \bullet OH_{1}.\]

\[2)\frac{\text{AO}}{OA_{1}} = \frac{\text{CO}}{OC_{1}} = \frac{\text{BO}}{OB_{1}} =\]

\[= \frac{2}{1}\ (по\ свойству\ медианы).\]

\[3)\frac{S_{\text{AOB}}}{S_{\text{OB}A_{1}}} = \frac{\text{AO}}{OA_{1}} = \frac{2}{1}\]

\[\frac{S}{S_{\text{OB}A_{1}}} = 2\]

\[S_{\text{OB}A_{1}} = \frac{S}{2}.\]

\[4)\frac{S_{\text{AOB}}}{S_{\text{AO}B_{1}}} = \frac{\text{BO}}{OB_{1}} = \frac{2}{1}\]

\[\frac{S}{S_{\text{AO}B_{1}}} = 2\]

\[S_{\text{AO}B_{1}} = \frac{S}{2}.\]

\[5)\ S_{\text{AB}A_{1}} = S_{AA_{1}C}\ \]

\[(как\ равновеликие).\]

\[6)\ S_{\text{AB}A_{1}} = S_{\text{AOB}} + S_{BA_{1}O} =\]

\[= S + \frac{S}{2} = \frac{3}{2}\text{S.}\]

\[7)\ S_{\text{ABC}} = S_{\text{AB}A_{1}} + S_{AA_{1}C} =\]

\[= \frac{3}{2}S + \frac{3}{2}S = 3S.\]

\[Ответ:S_{\text{ABC}} = 3S.\]