\[\boxed{\mathbf{1128.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - квадрат;\]
\[AB = a.\]
\[\mathbf{Найти:}\]
\[S_{ост} - ?\]
\[\mathbf{Решение.}\]
\[1)\ ABCD - квадрат \Longrightarrow\]
\[\Longrightarrow S_{\text{ABCD}} = a^{2}:\]
\[\angle A = \angle B = \angle C = \angle D = 90{^\circ}\ \]
\[(по\ свойству).\]
\[2)\ S_{1} = S_{2} = S_{3} = S_{4} =\]
\[= \frac{\pi R^{2}}{360{^\circ}} \bullet \alpha = \frac{\pi\left( \frac{a}{2} \right)^{2}}{360{^\circ}} \bullet 90{^\circ} = \frac{\pi a^{2}}{16}.\]
\[3)\ S_{закр} = S_{1} \bullet 4 = \frac{\pi a^{2}}{16} \bullet 4 = \frac{\pi a^{2}}{4}.\]
\[4)\ S_{ост} = S_{\text{ABCD}} - S_{закр} =\]
\[= a^{2} - \frac{\pi a^{2}}{4} = \frac{4a^{2} - \pi a^{2}}{4} =\]
\[= a^{2}\frac{4 - \pi}{4}.\]
\[Ответ:S_{ост} = a^{2}\frac{4 - \pi}{4}.\]
\[\boxed{\mathbf{1128.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - квадрат;\]
\[AC \cap DB = O.\]
\[\mathbf{Найти:}\]
\[угол\ между\ векторами.\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ \left( \widehat{\overrightarrow{\text{AB}};\overrightarrow{\text{AC}}} \right) = 45{^\circ};\]
\[\textbf{б)}\ \left( \widehat{\overrightarrow{\text{AB}};\overrightarrow{\text{DA}}} \right) = 90{^\circ};\]
\[\textbf{в)}\ \left( \widehat{\overrightarrow{\text{OA}};\overrightarrow{\text{OB}}} \right) = 90{^\circ};\]
\[\textbf{г)}\ \left( \widehat{\overrightarrow{\text{AO}};\overrightarrow{\text{OB}}} \right) = 90{^\circ};\]
\[\textbf{д)}\ \left( \widehat{\overrightarrow{\text{OA}};\overrightarrow{\text{OC}}} \right) = 180{^\circ};\]
\[\textbf{е)}\ \left( \widehat{\overrightarrow{\text{AC}};\overrightarrow{\text{BD}}} \right) = 90{^\circ};\]
\[\textbf{ж)}\ \left( \widehat{\overrightarrow{\text{AD}};\overrightarrow{\text{DB}}} \right) = 135{^\circ};\]
\[\textbf{з)}\ \left( \widehat{\overrightarrow{\text{AO}};\overrightarrow{\text{OC}}} \right) = 0{^\circ}.\]