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МатематикаРешебник по геометрии 9 класс Атанасян ФГОС Задание 784
Задание 784
\[\boxed{\mathbf{784.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\mathbf{\ задачи:}\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[BD \cap AC = 0;\]
\[AM = \frac{1}{2}MD;\]
\[\overrightarrow{x} = \overrightarrow{\text{AD}};\ \ \]
\[\overrightarrow{y} = \overrightarrow{\text{AB}}.\]
\[\mathbf{Выразить:}\]
\[\textbf{а)}\ \ \overrightarrow{\text{AC}};\ \overrightarrow{\text{AO}};\overrightarrow{\text{CO}};\overrightarrow{\text{DO}};\]
\[\overrightarrow{\text{AD}} + \overrightarrow{\text{BC}};\overrightarrow{\text{AD}} + \overrightarrow{\text{CO}};\overrightarrow{\text{CO}} + \overrightarrow{\text{OA}}.\]
\[\textbf{б)}\ \overrightarrow{\text{AM}};\overrightarrow{\text{MC}};\overrightarrow{\text{BM}};\overrightarrow{\text{OM}}.\]
\[\mathbf{Решение.}\]
\[ABCD - параллелограмм:\]
\[AB = CD\ \ и\ \ BC = AD;\]
\[BO = OD\ \ и\ \ AO = OC.\]
\[По\ условию:\ \]
\[AM = \frac{1}{2} \bullet MD \Longrightarrow\]
\[\Longrightarrow AM\ :MD = 1\ :2.\ \]
\[\textbf{а)}\ 1)\ \overrightarrow{\text{AC}} = \overrightarrow{\text{AD}} + \overrightarrow{\text{DC}} =\]
\[= \overrightarrow{\text{AD}} + \overrightarrow{\text{AB}} = \overrightarrow{x} + \overrightarrow{y}.\]
\[2)\ \overrightarrow{\text{AO}} = \frac{1}{2}\overrightarrow{\text{AC}} = \frac{1}{2} \bullet \left( \overrightarrow{x} + \overrightarrow{y} \right) =\]
\[= \frac{1}{2}\overrightarrow{x} + \frac{1}{2}\overrightarrow{y}.\]
\[3)\ \overrightarrow{\text{CO}} = - \overrightarrow{\text{OC}} = - \overrightarrow{\text{AO}} =\]
\[= - \left( \frac{1}{2}\overrightarrow{x} + \frac{1}{2}\overrightarrow{y} \right) = - \frac{1}{2}\overrightarrow{x} - \frac{1}{2}\overrightarrow{y}.\]
\[4)\ \overrightarrow{\text{DO}} = \overrightarrow{\text{DA}} + \overrightarrow{\text{AO}} =\]
\[= - \overrightarrow{\text{AD}} + \overrightarrow{\text{AO}} = - \overrightarrow{x} + \frac{1}{2}\overrightarrow{x} + \frac{1}{2}\overrightarrow{y} =\]
\[= - \frac{1}{2}\overrightarrow{x} + \frac{1}{2}\overrightarrow{y}.\]
\[5)\ \overrightarrow{\text{AD}} + \overrightarrow{\text{BC}} = \overrightarrow{\text{AD}} + \overrightarrow{\text{AD}} =\]
\[= 2\overrightarrow{\text{AD}} = 2\overrightarrow{x}.\]
\[6)\ \overrightarrow{\text{AD}} + \overrightarrow{\text{CO}} = \overrightarrow{x} - \frac{1}{2}\overrightarrow{x} - \frac{1}{2}\overrightarrow{y} =\]
\[= \frac{1}{2}\overrightarrow{x} - \frac{1}{2}\overrightarrow{y}.\]
\[7)\ \overrightarrow{\text{CO}} + \overrightarrow{\text{OA}} = \overrightarrow{\text{CA}} = - \overrightarrow{\text{AC}} =\]
\[= - \left( \overrightarrow{x} + \overrightarrow{y} \right) = - \overrightarrow{x} - \overrightarrow{y}.\]
\[\textbf{б)}\ 1)\ \overrightarrow{\text{AM}} = \frac{1}{3}\overrightarrow{\text{AD}} = \frac{1}{3}\overrightarrow{x}.\]
\[2)\ \overrightarrow{\text{MC}} = \overrightarrow{\text{MD}} + \overrightarrow{\text{DC}} =\]
\[= \frac{2}{3}\overrightarrow{\text{AD}} + \left( \overrightarrow{\text{AB}} \right) = \frac{2}{3}\overrightarrow{x} + \overrightarrow{y}.\]
\[3)\ \overrightarrow{\text{BM}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AM}} =\]
\[= - \overrightarrow{\text{AB}} + \overrightarrow{\text{AM}} = - \overrightarrow{\text{AB}} + \frac{1}{3}\overrightarrow{\text{AD}} =\]
\[= - \overrightarrow{y} + \frac{1}{3}\overrightarrow{x}.\]
\[4)\ \ \overrightarrow{\text{OM}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AM}} =\]
\[= - \overrightarrow{\text{AO}} + \frac{1}{3}\overrightarrow{\text{AD}} =\]
\[= - \frac{1}{2}\overrightarrow{x} - \frac{1}{2}\overrightarrow{y} + \frac{1}{3}\overrightarrow{x} = - \frac{1}{6}\overrightarrow{x} - \frac{1}{2}\overrightarrow{y}.\]
\[\boxed{\mathbf{784.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - четырехугольник;\]
\[четыреугол;\]
\[AB + CD = 15.\]
\[\mathbf{Найти:}\]
\[P_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ В\ \text{ABCD\ }можно\ вписать\ \]
\[окружность:\]
\[AB + CD = BC + AD\ \]
\[(по\ свойству\ вписанной\ \]
\[окружности\ в\ \ \]
\[четырехугольник).\]
\[2)\ P_{\text{ABCD}} =\]
\[= AB + BC + CD + AD =\]
\[= (AB + CD) \bullet 2 = 15 \bullet 2 =\]
\[= 30\ см.\]
\[Ответ:P_{\text{ABCD}} = 30\ см.\ \]
