Решебник по геометрии 7 класс Атанасян ФГОС Задание 1083

 

\[\boxed{\mathbf{1083.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Внутренние\ углы\ правильного\ \]

\[n - угольника\ находятся\ по\ \]

\[формуле:\]

\[\alpha = \frac{n - 2}{n} \bullet 180{^\circ}.\]

\[Отсюда:\ \]

\[\alpha n = (n - 2)180{^\circ}\]

\[\alpha n = 180{^\circ}n - 360{^\circ}\]

\[360{^\circ} = 180{^\circ}n - \alpha n\]

\[360{^\circ} = n(180{^\circ} - \alpha)\]

\[n = \frac{360{^\circ}}{180{^\circ} - \alpha}.\]

\[\textbf{а)}\ \alpha = 60{^\circ}:\ \ \ \]

\[n = \frac{360{^\circ}}{180{^\circ} - 60{^\circ}} = \frac{360{^\circ}}{120{^\circ}} = 3;\]

\[\textbf{б)}\ \alpha = 90{^\circ}:\ \ \ \]

\[n = \frac{360{^\circ}}{180{^\circ} - 90{^\circ}} = \frac{360{^\circ}}{90{^\circ}} = 4;\]

\[\textbf{в)}\ \alpha = 135{^\circ}:\ \ \ \]

\[n = \frac{360{^\circ}}{180{^\circ} - 135{^\circ}} = \frac{360{^\circ}}{45{^\circ}} = 8;\]

\[\textbf{г)}\ \alpha = 150{^\circ}:\ \ \ \]

\[n = \frac{360{^\circ}}{180{^\circ} - 150{^\circ}} = \frac{360{^\circ}}{30{^\circ}} = 12.\]

\[\boxed{\mathbf{1083.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

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

\[A( - 5;13);B(3;5);\]

\[C( - 3; - 1);\]

\[M,N,K - середины\ сторон.\]

\[\mathbf{Найти:}\]

\[\textbf{а)}\ координаты - \ M,N\ и\ K;\]

\[\textbf{б)}\ BK;\]

\[\textbf{в)}\ MN;MK;NK.\]

\[\mathbf{Решение.}\]

\[\textbf{б)}\ BK = \sqrt{(3 + 4)^{2} + (5 - 6)^{2}} =\]

\[= \sqrt{50} = 5\sqrt{2}.\]

\[\textbf{в)}\ 1)\ MN =\]

\[= \sqrt{( - 1 - 0)^{2} + (9 - 2)^{2}} =\]

\[= \sqrt{50} = 5\sqrt{2};\]

\[2)\ MK =\]

\[= \sqrt{( - 1 + 4)^{2} + (9 - 6)^{2}} =\]

\[= \sqrt{18} = 3\sqrt{2};\]

\[3)\ NK = \sqrt{(0 + 4)^{2} + (2 - 6)^{2}} =\]

\[= \sqrt{32} = 4\sqrt{2}.\]

\[Ответ:а)\ M( - 1;9);\ N(0;2);\ \]

\[K( - 4;6);\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ \ BK = \ 5\sqrt{2};\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ в)\ MN = 5\sqrt{2};\]

\(MK = 3\sqrt{2};NK = 4\sqrt{2}.\)