\[\boxed{\mathbf{1131.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[A_{1}A_{2}A_{3}A_{4}A_{5}A_{6} - правильный\ \]
\[шестиугольник;\]
\[A_{1}A_{4} = 2,24\ см.\]
\[\mathbf{Найти:}\]
\[P - ?\]
\[\mathbf{Решение.}\]
\[1)\ A_{1}A_{4} = d\]
\[2R = 2,24\ см\]
\[R = \frac{2,24}{2} = 1,12\ см.\]
\[2)\ a_{6} = 2R \bullet \sin\frac{180{^\circ}}{6} =\]
\[= 2 \bullet 1,12 \bullet \sin{30{^\circ}} = 2 \bullet 1,12 \bullet \frac{1}{2} =\]
\[= 1,12\ см.\]
\[3)\ P = 6 \bullet a_{6} = 6 \bullet 1,12 = 6,72\ см.\]
\[Ответ:P = 6,72\ см.\]
\[\boxed{\mathbf{1131.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - равносторон;\]
\[AB = a;\]
\[BD\bot AC.\]
\[\mathbf{Найти:}\]
\[скалярное\ произведение\ \]
\[векторов.\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}ABC - равносторонний:\ \]
\[\angle A = \angle B = \angle C =\]
\[= 60{^\circ}\ (по\ свойству).\]
\[2)\ BD\bot AC \Longrightarrow \angle ADB = 90{^\circ}.\]
\[\textbf{а)}\ \overrightarrow{\text{AB}} \bullet \overrightarrow{\text{AC}} = a \bullet a \bullet \cos{60{^\circ}} =\]
\[= a^{2} \bullet \frac{1}{2} = \frac{a^{2}}{2};\]
\[\textbf{б)}\ \overrightarrow{\text{AC}} \bullet \overrightarrow{\text{CB}} = a \bullet a \bullet \cos{120{^\circ}} =\]
\[= a^{2} \bullet \left( - \frac{1}{2} \right) = - \frac{a^{2}}{2};\]
\[\textbf{в)}\ \overrightarrow{\text{AC}} \bullet \overrightarrow{\text{BD}} =\]
\[= \left| \overrightarrow{\text{AC}} \right| \bullet \left| \overrightarrow{\text{BD}} \right| \bullet \cos{90{^\circ}} = 0;\]
\[\textbf{г)}\ \overrightarrow{\text{AC}} \bullet \overrightarrow{\text{AC}} = a \bullet a \bullet \cos{0{^\circ}} =\]
\[= a^{2} \bullet 1 = a^{2}.\]