Решебник по геометрии 9 класс Атанасян ФГОС Задание 1130

 

\[\boxed{\mathbf{1130.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

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

\[ACDE - квадрат;\]

\[R_{\mathrm{\Delta}} = 3\ дм.\]

\[\mathbf{Найти:}\]

\[R_{кв} - ?\]

\[\mathbf{Решение.}\]

\[1)\ a_{3} = 2R \bullet \sin\frac{180{^\circ}}{3} =\]

\[= 2 \bullet 3 \bullet \sin{60{^\circ}} = 6 \bullet \frac{\sqrt{3}}{2} =\]

\[= 3\sqrt{3}\ дм.\]

\[2)\ a_{4} = 3\sqrt{3}\ дм.\]

\[3)\ a_{4} = 2R \bullet \sin\frac{180{^\circ}}{4} =\]

\[= 2R \bullet \sin{45{^\circ}} = 2R \bullet \frac{\sqrt{2}}{2} = R\sqrt{2}.\]

\[4)\ 3\sqrt{3} = R\sqrt{2}\]

\[R = \frac{3\sqrt{3}}{\sqrt{2}} = \frac{3\sqrt{6}}{2}\ дм.\]

\[Ответ:R_{кв} = \frac{3\sqrt{6}}{2}\ дм.\]

\[\boxed{\mathbf{1130.еуроки - ответы\ на\ пятёрку}}\]

\[\mathbf{Дано:}\]

\[\left| \overrightarrow{a} \right| = 2;\left| \overrightarrow{b} \right| = 3;\]

\[\textbf{а)}\ 45{^\circ};\]

\[\textbf{б)}\ 90{^\circ};\]

\[\textbf{в)}\ 135{^\circ}.\]

\[\mathbf{Найти:}\]

\[\overrightarrow{a} \bullet \overrightarrow{b} - ?\]

\[\mathbf{Решение.}\]

\[\ \overrightarrow{a} \bullet \overrightarrow{b} = \left| \overrightarrow{a} \right| \bullet \left| \overrightarrow{b} \right| \bullet \cos{(\widehat{\overrightarrow{a};\overrightarrow{b}}}).\]

\[\textbf{а)}\ \overrightarrow{a} \bullet \overrightarrow{b} = 2 \bullet 3 \bullet \cos{45{^\circ}} =\]

\[= 6 \bullet \frac{\sqrt{2}}{2} = 3\sqrt{2}.\]

\[\textbf{б)}\ \overrightarrow{a} \bullet \overrightarrow{b} = 2 \bullet 3 \bullet \cos{90{^\circ}} =\]

\[= 6 \bullet 0 = 0.\]

\[\textbf{в)}\ \overrightarrow{a} \bullet \overrightarrow{b} = 2 \bullet 3 \bullet \cos{135{^\circ}} =\]

\[= 6 \bullet \left( - \frac{\sqrt{2}}{2} \right) = - 3\sqrt{2}.\]