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МатематикаРешебник по геометрии 7 класс Атанасян ФГОС Задание 763
Задание 763
\[\boxed{\mathbf{763.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[\mathrm{\Delta}ABC;\]
\[AB = 6;\]
\[BC = 8;\ \]
\[\angle B = 90{^\circ}.\]
\[Решение.\]
\[\textbf{а)}\ \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{BA}} \right| - \left| \overrightarrow{\text{BC}} \right| = 6 - 8 = - 2;\]
\[\left| \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} \right| =\]
\[= \left| \overrightarrow{\text{CA}} \right|\ (по\ правилу\ треугольника);\]
\[CA = \sqrt{AB^{2} + BC^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\ \]
\[\left| \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} \right| = 10.\]
\[\textbf{б)}\ \left| \overrightarrow{\text{AB}} \right| = \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{AB}} \right| + \left| \overrightarrow{\text{BC}} \right| = 6 + 8 = 14;\]
\[\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} \right| =\]
\[= \left| \overrightarrow{\text{AC}} \right|\ (по\ правилу\ треугольника);\]
\[AC = \sqrt{AB^{2} + BC^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\ \]
\[\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} \right| = 10.\]
\[\textbf{в)}\ \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{BA}} \right| + \left| \overrightarrow{\text{BC}} \right| = 6 + 8 = 14;\]
\[BD = \sqrt{AB^{2} + AD^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\]
\[\left| \overrightarrow{\text{BA}} + \overrightarrow{\text{BC}} \right| = 10.\]
\[\textbf{г)}\ \left| \overrightarrow{\text{AB}} \right| = \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{AB}} \right| - \left| \overrightarrow{\text{BC}} \right| = 6 - 8 = - 2;\]
\[\left| \overrightarrow{\text{AB}} - \overrightarrow{\text{BC}} \right| = \left| \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} \right| =\]
\[= |\overrightarrow{\text{DB}}|\ (по\ правилу\ треугольника);\]
\[DB = \sqrt{AB^{2} + AD^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\]
\[\left| \overrightarrow{\text{AB}} - \overrightarrow{\text{BC}} \right| = 10.\ \]
\[\boxed{\mathbf{763.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[окружность\ (O;r);\]
\[r = 16;\]
\[AB - хорда.\]
\[Найти:\]
\[AB - ?\]
\[Решение.\]
\[\textbf{а)}\ \angle AOB = 60{^\circ};\]
\[OA = OB = r = 16;\]
\[\mathrm{\Delta}ABO - равнобедренный\ \]
\[(по\ определению).\]
\[Отсюда:\ \]
\[\angle B = \angle A = \frac{180{^\circ} - 60{^\circ}}{2} = 60{^\circ};\]
\[\mathrm{\Delta}ABO - равносторонний \Longrightarrow\]
\[\Longrightarrow OA = OB = AB = 16.\]
\[\textbf{б)}\ \angle AOB = 90{^\circ};\]
\[OA = OB = r = 16;\ \]
\[\mathrm{\Delta}ABO - прямоугольный.\]
\[По\ теореме\ Пифагора:\]
\[AB = \sqrt{AO^{2} + OB^{2}} =\]
\[= \sqrt{16^{2} + 16^{2}} = \sqrt{2 \bullet 16^{2}} =\]
\[= 16\sqrt{2}\text{\ .}\]
\[\textbf{в)}\ \angle AOB = 180{^\circ};\]
\[OA = OB = r = 16;\ \]
\[\angle AOB - развернутый.\]
\[Следовательно:\]
\[AB = OA + OB = 2 \bullet 16 = 32.\]
\[Ответ:а)\ 16;б)\ 16\sqrt{2};в)\ 32.\]
