ГДЗ по геометрии 9 класс Атанасян ФГОС Задание 937

 

\[\boxed{\mathbf{937.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\mathbf{Дано:}\]

\[\text{A\ }(0;1);\]

\[\text{B\ }(5; - 3);\]

\[AB = BC;\]

\[BD = DC;\]

\[B \in AC;\ \]

\[D \in BC.\]

\[\mathbf{Найти:}\]

\[координаты\ C\ и\ \text{D.}\]

\[\mathbf{Решение.}\]

\[1)\ \left\{ \begin{matrix} x_{B} = \frac{x_{A} + x_{C}}{2} \\ y_{B} = \frac{y_{A} + y_{C}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} 5 = \frac{0 + x_{C}}{2}\text{\ \ \ \ } \\ - 3 = \frac{1 + y_{C}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \frac{x_{C}}{2} = 5\ \ \ \ \ \ \ \\ \frac{y_{C}}{2} = - 3,5 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x_{C} = 10\ \\ y_{C} = - 7 \\ \end{matrix} \Longrightarrow C\ (10; - 7). \right.\ \]

\[2)\ \left\{ \begin{matrix} x_{D} = \frac{x_{B} + x_{C}}{2} \\ y_{D} = \frac{y_{B} + y_{c}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x_{D} = \frac{5 + 10}{2}\text{\ \ \ \ \ \ \ \ } \\ y_{D} = \frac{- 3 + ( - 7)}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x_{D} = 7,5 \\ y_{D} = - 5 \\ \end{matrix}\ \Longrightarrow D\ (7,5;\ - 5). \right.\ \]

\(Ответ:C\ (10; - 7);D\ (7,5; - 5).\)

\[\boxed{\mathbf{937.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ задачи:\]

\[Дано:\]

\[ABCD - равнобедренная\ \]

\[трапеция;\]

\[\text{S\ }и\ T - середины\ боковых\ \]

\[сторон.\]

\[Равны\ ли\ векторы\]

\[\textbf{а)}\ \overrightarrow{\text{NL}}\ и\ \overrightarrow{\text{KL}}:\]

\[\overrightarrow{\text{NL}}\ и\ \overrightarrow{\text{KL}} - не\ коллинеарны;\]

\[\overrightarrow{\text{NL}} \neq \overrightarrow{\text{KL}}.\]

\[\textbf{б)}\ \overrightarrow{\text{MS}}\ и\ \overrightarrow{\text{SN}}:\]

\[\overrightarrow{\text{MS}} \uparrow \uparrow \overrightarrow{\text{SN}}\ и\ \left| \overrightarrow{\text{MS}} \right| =\]

\[= \left| \overrightarrow{\text{SN}} \right|\ \left( так\ как\ S - середина\ \text{MN} \right);\]

\[\ \overrightarrow{\text{MS}} = \overrightarrow{\text{SN}}.\]

\[\textbf{в)}\ \overrightarrow{\text{MN}}\ и\ \overrightarrow{\text{KL}}:\]

\[\overrightarrow{\text{MN}}\ и\ \overrightarrow{\text{KL}} - не\ коллинеарны;\]

\[\ \overrightarrow{\text{MN}} \neq \overrightarrow{\text{KL}}.\]

\[\textbf{г)}\ \overrightarrow{\text{TS}}\ и\ \overrightarrow{\text{KM}}:\]

\[\overrightarrow{\text{TS}} \uparrow \uparrow \overrightarrow{\text{KM}}\ и\ \left| \overrightarrow{\text{TS}} \right| \neq \left| \overrightarrow{\text{KM}} \right|;\]

\[\ \overrightarrow{\text{TS}} \neq \overrightarrow{\text{KM}}.\]

\[\textbf{д)}\ \overrightarrow{\text{TL}}\ и\ \overrightarrow{\text{KT}}:\]

\[\overrightarrow{\text{TL}} \uparrow \uparrow \overrightarrow{\text{KT}}\ и\ \left| \overrightarrow{\text{TL}} \right| =\]

\[= \left| \overrightarrow{\text{KT}} \right|\ \left( так\ как\ T - середина\ \text{KL} \right);\ \]

\[\overrightarrow{\text{TL}} = \overrightarrow{\text{KT}}.\]

\[Ответ:а)\ нет;б)\ да;в)\ нет;\]

\[\textbf{г)}\ нет;д)\ да.\]