\[\boxed{\mathbf{1132.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\textbf{а)}\ Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - правильный;\]
\[MNEF - квадрат;\]
\[P_{3} = P_{\text{ABC}};\ \]
\[P_{4} = P_{\text{MNEF}}.\]
\[\mathbf{Найти:}\]
\[P_{3}\ :P_{4} - ?\]
\[\mathbf{Решение.}\]
\[1)\ a_{3} = 2R \bullet \sin\frac{180{^\circ}}{3} =\]
\[= 2R \bullet \sin{60{^\circ}} = 2R \bullet \frac{\sqrt{3}}{2} = R\sqrt{3}.\]
\[2)\ P_{3} = a_{3} \bullet 3 = 3\sqrt{3}.\]
\[3)\ a_{4} = 2R \bullet \frac{180{^\circ}}{4} =\]
\[= 2R \bullet \sin{45{^\circ}} = 2R \bullet \frac{\sqrt{2}}{2} = R\sqrt{2}.\]
\[4)\ P_{4} = a_{4} \bullet 4 = 4\sqrt{2}\text{R.}\]
\[5)\frac{P_{3}}{P_{4}} = \frac{3\sqrt{3}R}{4\sqrt{2}R} = \frac{3\sqrt{3}}{4\sqrt{2}} = \frac{3\sqrt{6}}{8}.\]
\[Ответ:P_{3}:P_{4} = \frac{3\sqrt{6}}{8}.\]
\[\mathbf{б)\ Рисунок\ по\ условию\ задачи:}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - правильный;\]
\[MNEF - квадрат;\]
\[P_{3} = P_{\text{ABC}};\ \]
\[P_{4} = P_{\text{MNEF}}.\]
\[\mathbf{Найти:}\]
\[P_{3}\ :P_{4} - ?\]
\[\mathbf{Решение.}\]
\[1)\ a_{3} = 2R \bullet \sin\frac{180{^\circ}}{3} =\]
\[= 2R \bullet \sin{60{^\circ}} = 2R \bullet \frac{\sqrt{3}}{2} = R\sqrt{3}.\]
\[2)\ a_{4} = 2R \bullet \frac{180{^\circ}}{4} =\]
\[= 2R \bullet \sin{45{^\circ}} = 2R \bullet \frac{\sqrt{2}}{2} = R\sqrt{2}.\]
\[3)\ r =\]
\[= R \bullet \cos{\frac{180{^\circ}}{3} = R \bullet \cos{60{^\circ}}} =\]
\[= R \bullet \frac{1}{2} = \frac{R}{2};\]
\[R = 2r.\]
\[4)\ r = R \bullet \cos\frac{180{^\circ}}{4} =\]
\[= R \bullet \cos{45{^\circ}} = R\frac{\sqrt{2}}{2}\]
\[R = \frac{2r}{\sqrt{2}}.\]
\[5)\ P_{3} = a_{3} \bullet 3 = 3\sqrt{3}R = 6\sqrt{3}\text{\ r.}\]
\[6)\ P_{4} = a_{4} \bullet 4 = 4\sqrt{2}R =\]
\[= \frac{2r}{\sqrt{2}} \bullet \sqrt{2} \bullet 4 = 8\ r.\]
\[7)\ \frac{P_{3}}{P_{4}} = \frac{6\sqrt{3}r}{8r} = \frac{3\sqrt{3}}{4}.\]
\[Ответ:P_{3}:P_{4} = \frac{3\sqrt{3}}{4}.\]
\[\boxed{\mathbf{1132.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\left| \overrightarrow{P} \right| = 8;\]
\[\left| \overrightarrow{Q} \right| = 15;\]
\[\angle A = 120{^\circ}.\]
\[\mathbf{Найти:}\]
\[\left| \overrightarrow{F} \right| - ?\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}PAA_{1} - прямоугольный:\]
\[\angle PAA_{1} = 120{^\circ} - 90{^\circ} = 30{^\circ}.\]
\[По\ свойству\ прямоугольного\ \]
\[треугольника:\]
\[PA_{1} = \frac{1}{2}AP = \frac{8}{2} = 4.\]
\[2)\ \left\{ \begin{matrix} AA_{1} = \sqrt{AP^{2} - A_{1}P^{2}} \\ AA_{1} = \sqrt{AF^{2} - A_{1}F^{2}} \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \sqrt{AP^{2} - A_{1}P^{2}} =\]
\[= \sqrt{AF^{2} - A_{1}F^{2}};\]
\[8^{2} - 4^{2} = \overrightarrow{AF^{2}} - (15 - 4)\]
\[64 - 16 = \overrightarrow{AF^{2}} - 121\]
\[\overrightarrow{AF^{2}} = 64 + 121 - 16 = 169\]
\[\overrightarrow{\text{AF}} = \sqrt{169} = 13.\]
\[\mathbf{Ответ:}13\mathbf{.}\]