\[\boxed{\mathbf{698.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - четырехугольник;\]
\[четыреугольн;\]
\[r = 5\ см;\]
\[AB + CD = 12.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ В\ \text{ABCD\ }можно\ вписать\ \]
\[окружность:\]
\[AB + CD = BC + AD\ (по\ \]
\[свойству\ вписанной\ \]
\[окружности\ в\ \ \]
\[четырехугольник).\]
\[2)\ S_{\text{ABCD}} =\]
\[= S_{\text{AOB}} + S_{\text{BCO}} + S_{\text{COD}} + S_{\text{AOD}};\]
\[Ответ:S_{\text{ABCD}} = 60\ см^{2}.\ \]
\[\boxed{\mathbf{698.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ tg\ \alpha = \frac{1}{2}:\ \ \ \]
\[tg\ \alpha = \frac{\text{BC}}{\text{AC}} = \frac{1}{2}\]
\[BC = 1;\ AC = 2.\]
\[\textbf{б)}\ tg\ \alpha = \frac{3}{4}:\ \ \ \]
\[tg\ \alpha = \frac{\text{BC}}{\text{AC}} = \frac{3}{4}\]
\[BC = 3;\ AC = 4.\]
\[\textbf{в)}\cos\alpha = \frac{2}{10}:\]
\[\cos\alpha = \frac{\text{AC}}{\text{AB}} = \frac{1}{5}\]
\[AC = 1;\ AB = 5.\]
\[\textbf{г)}\cos\alpha = \frac{2}{3}:\]
\[\cos\alpha = \frac{\text{AC}}{\text{AB}} = \frac{2}{3}\]
\[\ AC = 2;\ AB = 3.\]
\[\textbf{д)}\sin\alpha = \frac{1}{2}:\]
\[\sin\alpha = \frac{\text{BC}}{\text{AB}} = \frac{1}{2}\]
\[BC = 1;\ AB = 2.\]
\[\textbf{е)}\sin\alpha = \frac{4}{10}:\ \ \]
\[\sin\alpha = \frac{\text{BC}}{\text{AB}} = \frac{2}{5}\]
\[BC = 2;\ AB = 5.\]