Решебник по геометрии 9 класс Атанасян ФГОС Задание 545

 

\[\boxed{\mathbf{545.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}\text{ABC}\sim\mathrm{\Delta}A_{1}B_{1}C_{1};\]

\[S_{\text{ABC}} = S_{A_{1}B_{1}C_{1}} + 77;\]

\[\frac{\text{AB}}{A_{1}B_{1}} = \frac{6}{5}.\]

\[\mathbf{Найти:}\]

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

\[S_{A_{1}B_{1}C_{1}} - ?\]

\[\mathbf{Решение.}\]

\[1)\ По\ теореме\ об\ отношении\ \]

\[площадей\ подобных\ \]

\[треугольников:\]

\[\frac{S_{\text{ABC}}}{S_{A_{1}B_{1}C_{1}}} = k^{2}\ \]

\[\frac{S_{\text{ABC}}}{S_{A_{1}B_{1}C_{1}}} = \left( \frac{6}{5} \right)^{2} = \frac{36}{25}\]

\[S_{\text{ABC}} = \frac{36}{25}S_{A_{1}B_{1}C_{1}}.\]

\[2)\ \frac{36}{25}S_{A_{1}B_{1}C_{1}} = S_{A_{1}B_{1}C_{1}} + 77\]

\[\frac{36}{25}S_{A_{1}B_{1}C_{1}} - S_{A_{1}B_{1}C_{1}} = 77\]

\[\frac{11}{25}S_{A_{1}B_{1}C_{1}} = 77\]

\[S_{A_{1}B_{1}C_{1}} = 77 \bullet \frac{25}{11} = 25 \bullet 7 =\]

\[= 175\ см^{2}.\]

\[3)\ S_{\text{ABC}} = 175 + 77 = 252\ см^{2}.\]

\[\mathbf{Ответ:}S_{\text{ABC}} = 252\ см^{2};\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }S_{A_{1}B_{1}C_{1}} = 175\ см^{2}\mathbf{.}\]

\[\boxed{\mathbf{545.еуроки - ответы\ на\ пятёрку}}\]

\[\textbf{а)}\ a = 1,2\ см:\ \]

\[S = (1,2)^{2} = 1,44\ см^{2}.\]

\[\textbf{б)}\ a = \frac{3}{4}\ дм:\]

\[S = \left( \frac{3}{4} \right)^{2} = \frac{9}{16}\ дм^{2}.\]

\[\textbf{в)}\ a = 3\sqrt{2}\ м:\]

\[S = \left( 3\sqrt{2} \right)^{2} = 9 \bullet 2 = 18\ м^{2}.\]