ГДЗ по геометрии 7 класс Атанасян Задание 950

\[\boxed{\mathbf{950.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[MNPQ - четырехугольник;\]

\[\textbf{а)}\ M(1;1);\]

\[N(6;1);\]

\[P(7;4);\]

\[Q(2;4).\]

\[\textbf{б)}\ M( - 5;1);\]

\[N( - 4;4);\]

\[P( - 1;5);\]

\[Q( - 2;2).\]

\[\mathbf{Доказать:}\]

\[MNPQ - параллелограмм.\]

\[Найти:\]

\[MN\ и\ \text{PQ.}\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ 1)\ MQ =\]

\[= \sqrt{(2 - 1)^{2} + (4 - 1)^{2}} =\]

\[= \sqrt{1 + 9} = \sqrt{10}\]

\[NP = \sqrt{(7 - 6)^{2} + (4 - 1)^{2}} =\]

\[= \sqrt{1 + 9} = \sqrt{10}\]

\[MN = \sqrt{(6 - 1)^{2} + (1 - 1)^{2}} =\]

\[= \sqrt{25 + 0} = 5\]

\[PQ = \sqrt{(2 - 7)^{2} + (4 - 4)^{2}} =\]

\[= \sqrt{25 + 0} = 5\]

\[3)\ NQ = \sqrt{(2 - 6)^{2} + (4 - 1)^{2}} =\]

\[= \sqrt{16 + 9} = 5;\]

\[MP = \sqrt{(7 - 1)^{2} + (4 - 1)^{2}} =\]

\[= \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}.\]

\[\textbf{б)}\ 1)\ MQ =\]

\[= \sqrt{( - 2 + 5)^{2} + (2 - 1)^{2}} =\]

\[= \sqrt{9 + 1} = \sqrt{10}\]

\[NP = \sqrt{( - 1 + 4)^{2} + (5 - 4)^{2}} =\]

\[= \sqrt{9 + 1} = \sqrt{10}\]

\[MN = \sqrt{( - 4 + 5)^{2} + (4 - 1)^{2}} =\]

\[= \sqrt{1 + 9} = \sqrt{10}\]

\[PQ = \sqrt{( - 2 + 1)^{2} + (2 - 5)^{2}} =\]

\[= \sqrt{1 + 9} = \sqrt{10}\]

\[3)\ NQ =\]

\[= \sqrt{( - 2 + 4)^{2} + (2 - 4)^{2}} =\]

\[= \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2};\]

\[MP = \sqrt{( - 1 + 5)^{2} + (5 - 1)^{2}} =\]

\[= \sqrt{16 + 16} = 4\sqrt{2}.\]

\[\mathbf{Ответ:}\mathbf{\ }а)\ NQ = 5;MP = 3\sqrt{2};\ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ NQ = 2\sqrt{2};MP = 4\sqrt{2}.\]