ГДЗ по геометрии 9 класс Атанасян Задание 762

\[\boxed{\mathbf{762.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ \mathbf{задачи:}\]

\[Дано:\ \]

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

\[AB = BC = AC = a.\]

\[Найти:\]

\[\textbf{а)}\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} \right|;\ \]

\[\textbf{б)}\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{AC}} \right|;\ \]

\[\textbf{в)}\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{CB}} \right|;\ \]

\[\textbf{г)}\left| \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} \right|;\ \]

\[\textbf{д)}\left| \overrightarrow{\text{AB}} - \overrightarrow{\text{AC}} \right|\text{.\ }\]

\[Решение.\]

\[\textbf{а)}\ \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} = \overrightarrow{\text{AC}} =\]

\[= a\ (по\ правилу\ треугольника).\]

\[\left| \overrightarrow{\text{AD}} \right| = 2AO;\]

\[AO = \sqrt{a^{2} - \left( \frac{a}{2} \right)^{2}} = \sqrt{\frac{3a^{2}}{4}} =\]

\[= \frac{a\sqrt{3}}{2}\]

\[\left| \overrightarrow{\text{AD}} \right| = 2 \bullet \frac{a\sqrt{3}}{2} = a\sqrt{3}.\]

\[\textbf{в)}\ \overrightarrow{\text{AB}} + \overrightarrow{\text{CB}} = \overrightarrow{\text{CD}} + \overrightarrow{\text{DE}} =\]

\[= \overrightarrow{\text{CE}}\ (DE \parallel BC;DE = BC):\]

\[CDEB - ромб\ \]

\[(по\ построению) \Longrightarrow CE = AD.\]

\[Следовательно:\]

\[\left| \overrightarrow{\text{CE}} \right| = a\sqrt{3}.\]

\[\textbf{г)}\ \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} = \overrightarrow{\text{CA}} =\]

\[= a\ (по\ правилу\ треугольника).\]

\[\textbf{д)}\ \overrightarrow{\text{AB}} - \overrightarrow{\text{AC}} = \overrightarrow{\text{CB}} =\]

\[= a\ (по\ правилу\ треугольника).\]

\[Ответ:а)\ a;б)\ a\sqrt{3};в)\ a\sqrt{3};\]

\[\textbf{г)}\ a;д)\ \text{a.}\]