\[\boxed{\mathbf{1136.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{Дано:}\]
\[A_{1}A_{2}A_{3}A_{4} - квадрат,\ \]
\[вписанный\ в\ окружность\ \]
\[радиуса\ \text{R.}\]
\[\mathbf{Доказать:}\]
\[B_{1}C_{3}B_{2}C_{4}B_{3}C_{1}B_{4}C_{2} -\]
\[правильный.\]
\[Выразить:\]
\[S_{8}\ через\ \text{R.}\]
\[\mathbf{Решение.}\]
\[1)\ A_{1}B_{1} = A_{2}C_{2} = R;\]
\[A_{1}A_{2} = A_{1}C_{2} + C_{2}B_{1} + B_{1}A_{2}.\]
\[если\ C_{2}B_{1} = x:\]
\[\ x + R - x + R - x = 2R - x =\]
\[= A_{1}A_{2}.\]
\[2)\ C_{3}B_{2} = C_{4}B_{3} = C_{1}B_{4} =\]
\[= 2R - A_{1}A_{2}:\]
\[A_{1}A_{2} = R\sqrt{2} \Longrightarrow \ C_{2}B_{1} = \ldots =\]
\[= B_{4}C_{1} = R\left( 2 - \sqrt{2} \right).\]
\[3)\ Докажем,\ что\ C_{2}B_{1} = B_{1}C_{3}.\]
\[Рассмотрим\ \mathrm{\Delta}A_{1}B_{1}C_{3}\text{.\ }\]
\[По\ теореме\ Пифагора:\]
\[B_{1}C_{3} = \sqrt{2(R - x)^{2}}\]
\[B_{1}C_{3} = \sqrt{2}(R - x)\]
\[B_{1}C_{3} = \sqrt{2}\left( R - 2R + R\sqrt{2} \right) =\]
\[= 2R - R\sqrt{2} = R\left( 2 - \sqrt{2} \right).\]
\[4)\ Получаем,\ что\ все\ стороны\ \]
\[восьмиугольника\ равны.\]
\[Значит,\ все\ его\ углы\ также\ \]
\[равны.\]
\[Получаем:\]
\[B_{1}C_{3}B_{2}C_{4}B_{3}C_{1}B_{4}C_{2} -\]
\[правильный\ многоугольник.\]
\[Что\ и\ требовалось\ доказать.\]
\[5)\ S = 8 \bullet S_{B_{1}OC_{2}};\ \ \]
\[O - точка\ пересечения\ \]
\[диагоналей:\]
\[S_{B_{1}OC_{2}} = \frac{1}{2}OB_{1} \bullet OC_{2} \bullet \sin{\angle B_{1}OC_{2}}.\]
\[7)\ \angle B_{1}OC_{2} = 45{^\circ}\ \]
\[(так\ как\ все\ углы\ \ по\ 135{^\circ}).\]
\[В\ \mathrm{\Delta}B_{1}OC_{2}:\ \]
\[\angle B_{1} = 67,5{^\circ};\ \]
\[\angle C_{2} = 67,5{^\circ}.\]
\[8)\ По\ теореме\ косинусов:\]
\[R^{2}\left( 2 - \sqrt{2} \right)^{2} =\]
\[= x^{2} + x^{2} - 2x^{2} \bullet \frac{\sqrt{2}}{2}\]
\[R^{2}\left( 2 - \sqrt{2} \right)^{2} = x^{2}\left( 2 - \sqrt{2} \right)\]
\[x = R\sqrt{2 - \sqrt{2}}.\]
\[8)\ S_{B_{1}OC_{2}} =\]
\[= \frac{1}{2}R\sqrt{2 - \sqrt{2}} \bullet R\sqrt{2 - \sqrt{2}} \bullet \sin{45{^\circ}} =\]
\[= \frac{1}{2}R^{2}\left( 2 - \sqrt{2} \right) \bullet \frac{\sqrt{2}}{2} =\]
\[= \frac{\left( \sqrt{2} - 1 \right)R^{2}}{2}.\]
\[9)\ S_{8} = 8 \bullet \frac{\left( \sqrt{2} - 1 \right)R^{2}}{2} =\]
\[= 4\left( \sqrt{2} - 1 \right)R^{2}.\]
\[Ответ:S_{8} = 4\left( \sqrt{2} - 1 \right)R^{2}.\]
\[\boxed{\mathbf{1136.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\ \]
\[\overrightarrow{a}\ и\ \overrightarrow{b}.\]
\[Найти:\]
\[x,\ при\ котором\ \overrightarrow{a}\bot\overrightarrow{b}.\]
\[Решение:\]
\[Если\ \overrightarrow{a}\bot\overrightarrow{b},\ то\ должно\ выполнять\ условие:\]
\[\overrightarrow{a} \bullet \overrightarrow{b} = 0.\]
\[\textbf{а)}\ \overrightarrow{a}\left\{ 4;5 \right\};\ \overrightarrow{b}\left\{ x; - 6 \right\}:\]
\[\overrightarrow{a} \bullet \overrightarrow{b} = 4x + 5 \bullet ( - 6) = 0\]
\[4x - 30 = 0\]
\[x = 7,5.\]
\[\textbf{б)}\ \overrightarrow{a}\left\{ x; - 1 \right\};\ \overrightarrow{b}\left\{ 3;2 \right\}:\]
\[\overrightarrow{a} \bullet \overrightarrow{b} = x \bullet 3 + ( - 1) \bullet 2 = 0\]
\[3x - 2 = 0\]
\[x = \frac{2}{3}.\]
\[\textbf{в)}\ \overrightarrow{a}\left\{ 0; - 3 \right\};\ \overrightarrow{b}\left\{ 5;x \right\}:\]
\[\overrightarrow{a} \bullet \overrightarrow{b} = 0 \bullet 5 + ( - 3)x = 0\]
\[- 3x = 0\]
\[x = 0.\]