\[\boxed{\mathbf{545.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC}\sim\mathrm{\Delta}A_{1}B_{1}C_{1};\]
\[S_{\text{ABC}} = S_{A_{1}B_{1}C_{1}} + 77;\]
\[\frac{\text{AB}}{A_{1}B_{1}} = \frac{6}{5}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}} - ?\]
\[S_{A_{1}B_{1}C_{1}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ По\ теореме\ об\ отношении\ \]
\[площадей\ подобных\ \]
\[треугольников:\]
\[\frac{S_{\text{ABC}}}{S_{A_{1}B_{1}C_{1}}} = k^{2}\ \]
\[\frac{S_{\text{ABC}}}{S_{A_{1}B_{1}C_{1}}} = \left( \frac{6}{5} \right)^{2} = \frac{36}{25}\]
\[S_{\text{ABC}} = \frac{36}{25}S_{A_{1}B_{1}C_{1}}.\]
\[2)\ \frac{36}{25}S_{A_{1}B_{1}C_{1}} = S_{A_{1}B_{1}C_{1}} + 77\]
\[\frac{36}{25}S_{A_{1}B_{1}C_{1}} - S_{A_{1}B_{1}C_{1}} = 77\]
\[\frac{11}{25}S_{A_{1}B_{1}C_{1}} = 77\]
\[S_{A_{1}B_{1}C_{1}} = 77 \bullet \frac{25}{11} = 25 \bullet 7 =\]
\[= 175\ см^{2}.\]
\[3)\ S_{\text{ABC}} = 175 + 77 = 252\ см^{2}.\]
\[\mathbf{Ответ:}S_{\text{ABC}} = 252\ см^{2};\]
\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }S_{A_{1}B_{1}C_{1}} = 175\ см^{2}\mathbf{.}\]
\[\boxed{\mathbf{545.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ a = 1,2\ см:\ \]
\[S = (1,2)^{2} = 1,44\ см^{2}.\]
\[\textbf{б)}\ a = \frac{3}{4}\ дм:\]
\[S = \left( \frac{3}{4} \right)^{2} = \frac{9}{16}\ дм^{2}.\]
\[\textbf{в)}\ a = 3\sqrt{2}\ м:\]
\[S = \left( 3\sqrt{2} \right)^{2} = 9 \bullet 2 = 18\ м^{2}.\]