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МатематикаРешебник по геометрии 8 класс Атанасян ФГОС Задание 951
Задание 951
\[\boxed{\mathbf{951.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - четырехугольник;\]
\[\textbf{а)}\ A( - 3; - 1);\]
\[B(1; - 1);\]
\[C(1; - 3);\]
\[D( - 3; - 3).\]
\[\textbf{б)}\ A(4;1);\]
\[B(3;5);\]
\[C( - 1;4);\]
\[D(0;0).\]
\[\mathbf{Доказать:}\]
\[ABCD - прямоугольник.\]
\[Найти:\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ AB =\]
\[= \sqrt{(1 + 3)^{2} + ( - 1 + 1)^{2}} =\]
\[= \sqrt{16} = 4\]
\[CD = \sqrt{( - 3 - 1)^{2} + ( - 1 + 1)^{2}} =\]
\[= \sqrt{16} = 4\]
\[BC = \sqrt{(1 - 1)^{2} + ( - 3 + 1)^{2}} =\]
\[= \sqrt{4} = 2\]
\[AD = \sqrt{( - 3 + 3)^{2} + ( - 3 + 1)^{2}} =\]
\[= \sqrt{4} = 2\]
\[2)\ ABCD - параллелограмм\ \]
\[(по\ второму\ признаку):\]
\[AB = CD;\ \ BC = AD.\]
\[3)\ AC =\]
\[= \sqrt{(1 + 3)^{2} + ( - 3 + 1)^{2}} =\]
\[= \sqrt{16 + 4} = 2\sqrt{5};\]
\[BD = \sqrt{( - 3 - 1)^{2} + ( - 3 + 1)^{2}} =\]
\[= \sqrt{16 + 4} = 2\sqrt{5}.\]
\[4)\ AC = BD;\ \ \]
\[ABCD - параллелограмм:\]
\[ABCD - прямоугольник\ \]
\[(по\ признаку\ прямоугольника).\]
\[5)\ S_{\text{ABCD}} = AB \bullet AD = 4 \bullet 2 = 8.\]
\[\textbf{б)}\ 1)\ AB =\]
\[= \sqrt{(3 - 4)^{2} + (5 - 1)^{2}} =\]
\[= \sqrt{1 + 16} = \sqrt{17}\]
\[CD = \sqrt{(0 + 1)^{2} + (0 - 4)^{2}} =\]
\[= \sqrt{1 + 16} = \sqrt{17}\]
\[BC = \sqrt{( - 1 - 3)^{2} + (4 - 5)^{2}} =\]
\[= \sqrt{16 + 1} = \sqrt{17}\]
\[AD = \sqrt{(0 - 4)^{2} + (0 - 1)^{2}} =\]
\[= \sqrt{16 + 1} = \sqrt{17}\]
\[2)\ ABCD - параллелограмм\ \]
\[(по\ второму\ признаку):\]
\[AB = CD;\ \ BC = AD.\]
\[3)\ AC = \sqrt{( - 1 - 4)^{2} + (4 - 1)^{2}} =\]
\[= \sqrt{25 + 9} = \sqrt{34}\]
\[BD = \sqrt{(0 - 3)^{2} + (0 - 5)^{2}} =\]
\[= \sqrt{9 + 25} = \sqrt{34}\]
\[4)\ AC = BD;\ \ \]
\[ABCD - параллелограмм:\]
\[ABCD - прямоугольник\ \]
\[(по\ признаку\ прямоугольника).\]
\[5)\ S_{\text{ABCD}} = AB \bullet AD =\]
\[= \sqrt{17} \bullet \sqrt{17} = 17.\]
\[\mathbf{Ответ:}\mathbf{\ }а)\ 8;\ \ б)\ 17.\]
\[\boxed{\mathbf{951.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[\mathrm{\Delta}ABC;\]
\[AB = 6;\]
\[BC = 8;\ \]
\[\angle B = 90{^\circ}.\]
\[Решение.\]
\[\textbf{а)}\ \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{BA}} \right| - \left| \overrightarrow{\text{BC}} \right| = 6 - 8 = - 2;\]
\[\left| \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} \right| =\]
\[= \left| \overrightarrow{\text{CA}} \right|\ (по\ правилу\ треугольника);\]
\[CA = \sqrt{AB^{2} + BC^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\ \]
\[\left| \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} \right| = 10.\]
\[\textbf{б)}\ \left| \overrightarrow{\text{AB}} \right| = \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{AB}} \right| + \left| \overrightarrow{\text{BC}} \right| = 6 + 8 = 14;\]
\[AC = \sqrt{AB^{2} + BC^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\ \]
\[\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} \right| = 10.\]
\[\textbf{в)}\ \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{BA}} \right| + \left| \overrightarrow{\text{BC}} \right| = 6 + 8 = 14;\]
\[BD = \sqrt{AB^{2} + AD^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\]
\[\left| \overrightarrow{\text{BA}} + \overrightarrow{\text{BC}} \right| = 10.\]
\[\textbf{г)}\ \left| \overrightarrow{\text{AB}} \right| = \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{AB}} \right| - \left| \overrightarrow{\text{BC}} \right| = 6 - 8 = - 2;\]
\[\left| \overrightarrow{\text{AB}} - \overrightarrow{\text{BC}} \right| = \left| \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} \right| =\]
\[= |\overrightarrow{\text{DB}}|\ (по\ правилу\ треугольника);\]
\[DB = \sqrt{AB^{2} + AD^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\]
\[\left| \overrightarrow{\text{AB}} - \overrightarrow{\text{BC}} \right| = 10.\ \]
