\[\boxed{\mathbf{652.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[окружность\ (O,\ R);\ \]
\[R = 15\ см;\]
\[\cup AC = 37{^\circ};\]
\[\cup BD = 23{^\circ}.\]
\[\mathbf{Найти:}\]
\[CD - ?\]
\[\mathbf{Решение.}\]
\[1)\ AB - полуокружность \Longrightarrow\]
\[\Longrightarrow \ \cup AB = 180{^\circ}.\]
\[2)\ \cup CD = \cup AB - \cup AC - \cup BD;\]
\[\cup CD = 180{^\circ} - 37{^\circ} - 23{^\circ} =\]
\[= 120{^\circ}.\]
\[3)\ Построим\ OE\bot CD.\]
\[4)\ \ \mathrm{\Delta}COD - равнобедренный:\]
\[OD = CO = R.\ \]
\[\angle EOD = \frac{1}{2}\angle COD = 60{^\circ};\]
\[\ CE = ED.\]
\[5)\sin{\angle EOD} = \frac{\text{ED}}{\text{OD}}\]
\[ED = OD \bullet \sin{\angle EOD} =\]
\[= 15 \bullet \sin{60{^\circ}} = 15 \bullet \frac{\sqrt{3}}{2}.\]
\[6)\ CD = 2ED = 15\sqrt{3}\ см.\]
\[Ответ:CD = 15\sqrt{3}\ см.\]
\[\boxed{\mathbf{652.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}A_{1}B_{1}C_{1};\]
\[S_{\text{ABC}} = S_{A_{1}B_{1}C_{1}} + 77;\]
\[\frac{\text{AB}}{A_{1}B_{1}} = \frac{6}{5}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}} - ?\]
\[S_{A_{1}B_{1}C_{1}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ По\ теореме\ об\ отношении\ \]
\[площадей\ подобных\ \]
\[треугольников:\]
\[\frac{S_{\text{ABC}}}{S_{A_{1}B_{1}C_{1}}} = k^{2}\ \]
\[\frac{S_{\text{ABC}}}{S_{A_{1}B_{1}C_{1}}} = \left( \frac{6}{5} \right)^{2} = \frac{36}{25}\]
\[S_{\text{ABC}} = \frac{36}{25}S_{A_{1}B_{1}C_{1}}.\]
\[2)\ \frac{36}{25}S_{A_{1}B_{1}C_{1}} = S_{A_{1}B_{1}C_{1}} + 77\]
\[\frac{36}{25}S_{A_{1}B_{1}C_{1}} - S_{A_{1}B_{1}C_{1}} = 77\]
\[\frac{11}{25}S_{A_{1}B_{1}C_{1}} = 77\]
\[S_{A_{1}B_{1}C_{1}} = 77 \bullet \frac{25}{11} = 25 \bullet 7 =\]
\[= 175\ см^{2}.\]
\[3)\ S_{\text{ABC}} = 175 + 77 = 252\ см^{2}.\]
\[\mathbf{Ответ:}S_{\text{ABC}} = 252\ см^{2};\]
\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }S_{A_{1}B_{1}C_{1}} = 175\ см^{2}\mathbf{.}\]