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МатематикаГДЗ по геометрии 9 класс Атанасян ФГОС Задание 959
Задание 959
\[\boxed{\mathbf{959.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\textbf{а)}\ x^{2} + y^{2} = 9\ \Longrightarrow \ O(0;0);\]
\[R = 3:\]
\[\textbf{б)}\ (x - 1)^{2} + (y + 2)^{2} = 4\ \Longrightarrow \ \ \]
\[\Longrightarrow O(1; - 2);R = 2:\]
\[\textbf{в)}\ (x + 5)^{2} + (y - 3)^{2} = 25\ \Longrightarrow\]
\[\Longrightarrow \ O( - 5;3);R = 5:\]
\[\textbf{г)}\ (x - 1)^{2} + y^{2} = 4\ \Longrightarrow O(1;0);\]
\[R = 2:\]
\[\textbf{д)}\ x^{2} + (y + 2)^{2} = 2\ \Longrightarrow\]
\[\Longrightarrow O(0; - 2);R = \sqrt{2}\mathbf{:\ }\]
\[\boxed{\mathbf{959.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[ABCD - параллелограмм;\ \]
\[AC \cap BD = O;\ \]
\[\overrightarrow{a} = \overrightarrow{\text{AB}};\ \]
\[\overrightarrow{b} = \overrightarrow{\text{AD}}.\]
\[Выразить:\ \]
\[\overrightarrow{\text{DC}} + \overrightarrow{\text{CB}};\ \]
\[\overrightarrow{\text{BO}} + \overrightarrow{\text{OC}};\ \]
\[\overrightarrow{\text{BO}} - \overrightarrow{\text{OC}};\ \]
\[\overrightarrow{\text{BA}} - \overrightarrow{\text{DA}}.\]
\[Решение.\]
\[1)\ \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} = \overrightarrow{\text{DB}} =\]
\[= \overrightarrow{\text{AB}} - \overrightarrow{\text{AD}} =\]
\[= \overrightarrow{a} - \overrightarrow{b}\ (по\ правилу\ треугольника).\]
\[2)\ \overrightarrow{\text{BO}} + \overrightarrow{\text{OC}} = \overrightarrow{\text{BC}} = \overrightarrow{\text{AD}} = \overrightarrow{b}.\]
\[3)\ \overrightarrow{\text{BO}} - \overrightarrow{\text{OC}} = \overrightarrow{\text{BO}} - \overrightarrow{\text{AO}} =\]
\[= \overrightarrow{\text{BO}} - \left( - \overrightarrow{\text{OA}} \right) = \overrightarrow{\text{BO}} + \overrightarrow{\text{OA}} =\]
\[= \overrightarrow{\text{BA}} = - \overrightarrow{\text{AB}} =\]
\[= - \overrightarrow{a}\ (по\ правилу\ треугольника).\]
\[4)\ \overrightarrow{\text{BA}} - \overrightarrow{\text{DA}} = \overrightarrow{\text{BA}} - \left( - \overrightarrow{\text{AD}} \right) =\]
\[= \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}} = - \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} =\]
\[= - \overrightarrow{a} + \overrightarrow{b} = \overrightarrow{b} - \overrightarrow{a}.\]
