\[\boxed{\mathbf{1130.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - равносторонний;\]
\[ACDE - квадрат;\]
\[R_{\mathrm{\Delta}} = 3\ дм.\]
\[\mathbf{Найти:}\]
\[R_{кв} - ?\]
\[\mathbf{Решение.}\]
\[1)\ a_{3} = 2R \bullet \sin\frac{180{^\circ}}{3} =\]
\[= 2 \bullet 3 \bullet \sin{60{^\circ}} = 6 \bullet \frac{\sqrt{3}}{2} =\]
\[= 3\sqrt{3}\ дм.\]
\[2)\ a_{4} = 3\sqrt{3}\ дм.\]
\[3)\ a_{4} = 2R \bullet \sin\frac{180{^\circ}}{4} =\]
\[= 2R \bullet \sin{45{^\circ}} = 2R \bullet \frac{\sqrt{2}}{2} = R\sqrt{2}.\]
\[4)\ 3\sqrt{3} = R\sqrt{2}\]
\[R = \frac{3\sqrt{3}}{\sqrt{2}} = \frac{3\sqrt{6}}{2}\ дм.\]
\[Ответ:R_{кв} = \frac{3\sqrt{6}}{2}\ дм.\]
\[\boxed{\mathbf{1130.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\left| \overrightarrow{a} \right| = 2;\left| \overrightarrow{b} \right| = 3;\]
\[\textbf{а)}\ 45{^\circ};\]
\[\textbf{б)}\ 90{^\circ};\]
\[\textbf{в)}\ 135{^\circ}.\]
\[\mathbf{Найти:}\]
\[\overrightarrow{a} \bullet \overrightarrow{b} - ?\]
\[\mathbf{Решение.}\]
\[\ \overrightarrow{a} \bullet \overrightarrow{b} = \left| \overrightarrow{a} \right| \bullet \left| \overrightarrow{b} \right| \bullet \cos{(\widehat{\overrightarrow{a};\overrightarrow{b}}}).\]
\[\textbf{а)}\ \overrightarrow{a} \bullet \overrightarrow{b} = 2 \bullet 3 \bullet \cos{45{^\circ}} =\]
\[= 6 \bullet \frac{\sqrt{2}}{2} = 3\sqrt{2}.\]
\[\textbf{б)}\ \overrightarrow{a} \bullet \overrightarrow{b} = 2 \bullet 3 \bullet \cos{90{^\circ}} =\]
\[= 6 \bullet 0 = 0.\]
\[\textbf{в)}\ \overrightarrow{a} \bullet \overrightarrow{b} = 2 \bullet 3 \bullet \cos{135{^\circ}} =\]
\[= 6 \bullet \left( - \frac{\sqrt{2}}{2} \right) = - 3\sqrt{2}.\]