ГДЗ по геометрии 7 класс Атанасян Задание 763

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Год:2020-2021-2022
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Задание 763

\[\boxed{\mathbf{763.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[Дано:\ \]

\[\mathrm{\Delta}ABC;\]

\[AB = 6;\]

\[BC = 8;\ \]

\[\angle B = 90{^\circ}.\]

\[Решение.\]

\[\textbf{а)}\ \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]

\[\left| \overrightarrow{\text{BA}} \right| - \left| \overrightarrow{\text{BC}} \right| = 6 - 8 = - 2;\]

\[\left| \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} \right| =\]

\[= \left| \overrightarrow{\text{CA}} \right|\ (по\ правилу\ треугольника);\]

\[CA = \sqrt{AB^{2} + BC^{2}} =\]

\[= \sqrt{36 + 64} = \sqrt{100} = 10;\ \]

\[\left| \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} \right| = 10.\]

\[\textbf{б)}\ \left| \overrightarrow{\text{AB}} \right| = \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]

\[\left| \overrightarrow{\text{AB}} \right| + \left| \overrightarrow{\text{BC}} \right| = 6 + 8 = 14;\]

\[\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} \right| =\]

\[= \left| \overrightarrow{\text{AC}} \right|\ (по\ правилу\ треугольника);\]

\[AC = \sqrt{AB^{2} + BC^{2}} =\]

\[= \sqrt{36 + 64} = \sqrt{100} = 10;\ \]

\[\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} \right| = 10.\]

\[\textbf{в)}\ \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]

\[\left| \overrightarrow{\text{BA}} \right| + \left| \overrightarrow{\text{BC}} \right| = 6 + 8 = 14;\]

\[BD = \sqrt{AB^{2} + AD^{2}} =\]

\[= \sqrt{36 + 64} = \sqrt{100} = 10;\]

\[\left| \overrightarrow{\text{BA}} + \overrightarrow{\text{BC}} \right| = 10.\]

\[\textbf{г)}\ \left| \overrightarrow{\text{AB}} \right| = \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]

\[\left| \overrightarrow{\text{AB}} \right| - \left| \overrightarrow{\text{BC}} \right| = 6 - 8 = - 2;\]

\[\left| \overrightarrow{\text{AB}} - \overrightarrow{\text{BC}} \right| = \left| \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} \right| =\]

\[= |\overrightarrow{\text{DB}}|\ (по\ правилу\ треугольника);\]

\[DB = \sqrt{AB^{2} + AD^{2}} =\]

\[= \sqrt{36 + 64} = \sqrt{100} = 10;\]

\[\left| \overrightarrow{\text{AB}} - \overrightarrow{\text{BC}} \right| = 10.\ \]

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