\[\boxed{\mathbf{955.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AE - медиана;\]
\[BH = 10\ см;\]
\[AH = 10\ см;\]
\[HC = 4\ см.\]
\[\mathbf{Найти:}\]
\[AE - ?\]
\[\mathbf{Решение.}\]
\[1)\ Прямоугольная\ система\ \]
\[координат:\]
\[B(0;10);A( - 10;0);C(4;0);\]
\[H(0;0).\]
\[2)\ AE - медиана \Longrightarrow BE = EC.\]
\[3)\ \left\{ \begin{matrix} x_{E} = \frac{x_{C} + x_{B}}{2} \\ y_{E} = \frac{y_{C} + y_{B}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x_{E} = \frac{4 + 0}{2}\text{\ \ } \\ y_{E} = \frac{0 + 10}{2} \\ \end{matrix} \right.\ \Longrightarrow E(2;5).\]
\[4)\ AE =\]
\[= \sqrt{(2 + 10)^{2} + (5 - 0)^{2}} =\]
\[= \sqrt{12^{2} + 5^{2}} = \sqrt{144 + 25} =\]
\[= \sqrt{169} = 13\ см.\]
\[\mathbf{Ответ:}\mathbf{\ }13\ см.\]
\[\boxed{\mathbf{955.еуроки - ответы\ на\ пятёрку}}\]
\[Дано:\ \]
\[\mathrm{\Delta}ABC;\ \]
\[\overrightarrow{a} = \overrightarrow{\text{AB}};\ \]
\[\overrightarrow{b} = \overrightarrow{\text{AC}}.\]
\[Выразить:\]
\[\textbf{а)}\ \overrightarrow{\text{BA}};\]
\[\textbf{б)}\ \overrightarrow{\text{CB}};\]
\[\textbf{в)}\ \overrightarrow{\text{CB}} + \overrightarrow{\text{BA}};\]
\[Решение.\]
\[\textbf{а)}\ \overrightarrow{\text{BA}}\ и\ \overrightarrow{\text{AB}} -\]
\[противоположные:\]
\[\overrightarrow{\text{BA}} = - \overrightarrow{\text{AB}}\]
\[\overrightarrow{\text{BA}} = - \overrightarrow{a}.\]
\[\overrightarrow{\text{CA}} = - \overrightarrow{\text{AC}};\]
\[\overrightarrow{\text{CB}} = \overrightarrow{\text{AB}} - \overrightarrow{\text{AC}} = \overrightarrow{a} - \overrightarrow{b}.\]
\[\overrightarrow{\text{CA}} = - \overrightarrow{\text{AC}} = - \overrightarrow{b}.\]