\[\boxed{\mathbf{947.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{а)}\]
\[\mathbf{б)}\]
\[\mathbf{Дано:}\]
\[\textbf{а)}\ A(0;1);B(1; - 4);\ C(5;2);\]
\[\textbf{б)}A( - 4;1);B( - 2;4);C(0;1).\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}}\mathbf{- ?}\]
\[\mathbf{Доказать:}\]
\[\mathrm{\Delta}ABC - равнобедренный.\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ AB =\]
\[= \sqrt{(1 - 0)^{2} + ( - 4 - 1)^{2}} =\]
\[= \sqrt{1 + 25} = \sqrt{26}\]
\[BC = \sqrt{(5 - 1)^{2} + (2 + 4)^{2}} =\]
\[= \sqrt{16 + 36} = \sqrt{52} = \sqrt{4 \bullet 13} =\]
\[= 2\sqrt{13}\]
\[AC = \sqrt{(5 - 0)^{2} + (2 - 1)^{2}} =\]
\[= \sqrt{25 + 1} = \sqrt{26}\]
\[AB = AC = \sqrt{26} \Longrightarrow\]
\[\Longrightarrow \mathrm{\Delta}ABC - равнобедренный.\]
\[Что\ и\ требовалось\ доказать.\]
\[\left\{ \begin{matrix} x_{M} = \frac{x_{B} + x_{C}}{2} \\ y_{M} = \frac{y_{B} + y_{C}}{2} \\ \end{matrix}\text{\ \ \ } \right.\ \]
\[\left\{ \begin{matrix} x_{M} = \frac{1 + 5}{2}\text{\ \ \ } \\ y_{M} = \frac{- 4 + 2}{2} \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x_{M} = 3\ \ \ \\ y_{M} = - 1 \\ \end{matrix}\ \Longrightarrow M(3; - 1). \right.\ \]
\[4)\ AM =\]
\[= \sqrt{(3 - 0)^{2} + ( - 1 - 1)^{2}} =\]
\[= \sqrt{9 + 4} = \sqrt{13}.\]
\[5)\ S_{\text{ABC}} = \frac{1}{2} \bullet AM \bullet BC =\]
\[= \frac{1}{2} \bullet \sqrt{13} \bullet 2\sqrt{13} = 13.\]
\[\textbf{б)}\ 1)\ AB =\]
\[= \sqrt{( - 2 + 4)^{2} + (4 - 1)^{2}} =\]
\[= \sqrt{4 + 9} = \sqrt{13}\]
\[BC = \sqrt{(0 + 2)^{2} + (1 - 4)^{2}} =\]
\[= \sqrt{4 + 9} = \sqrt{13}\]
\[AC = \sqrt{(0 + 4)^{2} + (1 - 1)^{2}} =\]
\[= \sqrt{16 + 0} = 4\]
\[AB = BC = \sqrt{13} \Longrightarrow\]
\[\Longrightarrow \mathrm{\Delta}ABC - равнобедренный.\]
\[Что\ и\ требовалось\ доказать.\]
\[\left\{ \begin{matrix} x_{M} = \frac{x_{A} + x_{C}}{2} \\ y_{M} = \frac{y_{A} + y_{C}}{2} \\ \end{matrix}\text{\ \ \ } \right.\ \]
\[\left\{ \begin{matrix} x_{M} = \frac{- 4 + 0}{2} \\ y_{M} = \frac{1 + 1}{2}\text{\ \ \ } \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x_{M} = - 2 \\ y_{M} = 1\ \ \ \\ \end{matrix}\ \Longrightarrow M( - 2;1) \right.\ .\]
\[3)\ BM =\]
\[= \sqrt{( - 2 + 2)^{2} + (1 - 4)^{2}} =\]
\[= \sqrt{0 + 9} = 3.\]
\[4)\ S_{\text{ABC}} = \frac{1}{2} \bullet BM \bullet AC =\]
\[= \frac{1}{2} \bullet 3 \bullet 4 = 6.\]
\[\mathbf{Ответ:}\mathbf{\ }а)\ 13;\ \ б)\ 6.\]
\[\boxed{\mathbf{947.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[MNPQ - четырехугольник.\]
\[\mathbf{Доказать:}\]
\[\textbf{а)}\ \overrightarrow{\text{MN}} + \overrightarrow{\text{NQ}} = \overrightarrow{\text{MP}} + \overrightarrow{\text{PQ}};\]
\[\textbf{б)}\ \overrightarrow{\text{MN}} + \overrightarrow{\text{NP}} = \overrightarrow{\text{MQ}} + \overrightarrow{\text{QP}}.\]
\[\mathbf{Доказательство.}\]
\[\textbf{а)}\ По\ правилу\ треугольника:\]
\[\overrightarrow{\text{MN}} + \overrightarrow{\text{NQ}} = \overrightarrow{\text{MQ}}\ и\ \overrightarrow{\text{MP}} + \overrightarrow{\text{PQ}} =\]
\[= \overrightarrow{\text{MQ}}.\]
\[Следовательно:\]
\[\overrightarrow{\text{MN}} + \overrightarrow{\text{NQ}} = \overrightarrow{\text{MP}} + \overrightarrow{\text{PQ}}.\]
\[Что\ и\ требовалось\ доказать.\]
\[\ б)\ По\ правилу\ треугольника:\]
\[\overrightarrow{\text{MN}} + \overrightarrow{\text{NP}} = \overrightarrow{\text{MP}}\ и\ \overrightarrow{\text{MQ}} + \overrightarrow{\text{QP}} =\]
\[= \overrightarrow{\text{MP}}.\]
\[Следовательно:\]
\[\overrightarrow{\text{MN}} + \overrightarrow{\text{NP}} = \overrightarrow{\text{MQ}} + \overrightarrow{\text{QP}}.\]
\[Что\ и\ требовалось\ доказать.\]