\[\boxed{\mathbf{1081.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Углы\ правильного\ \]
\[n - угольника\ находятся\ по\ \]
\[формуле:\ \]
\[\alpha = \frac{n - 2}{n} \bullet 180{^\circ}.\]
\[\textbf{а)}\ n = 3:\ \ \ \]
\[\alpha = \frac{3 - 2}{3} \bullet 180{^\circ} = 60{^\circ};\]
\[\textbf{б)}\ n = 5:\ \ \ \]
\[\alpha = \frac{5 - 2}{5} \bullet 180{^\circ} = 108{^\circ};\]
\[\textbf{в)}\ n = 6:\ \ \ \]
\[\alpha = \frac{6 - 2}{6} \bullet 180{^\circ} = 120{^\circ};\]
\[\textbf{г)}\ n = 10:\ \ \ \]
\[\alpha = \frac{10 - 2}{10} \bullet 180{^\circ} = 144{^\circ};\]
\[\textbf{д)}\ n = 18:\ \ \ \]
\[\alpha = \frac{18 - 2}{18} \bullet 180{^\circ} = 160{^\circ}.\]
\[\boxed{\mathbf{1081.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\textbf{а)}\ D(1;1);A(5;4);\]
\[B(4; - 3);C( - 2;5).\]
\[\textbf{б)}\ D(1;0);A(7; - 8);\]
\[B( - 5;8);C(9;6).\]
\[\mathbf{Доказать:}\]
\[AD = DB = DC.\]
\[\mathbf{Доказательство.}\]
\[\textbf{а)}\ 1)\ AD =\]
\[= \sqrt{(5 - 1)^{2} + (4 - 1)^{2}} =\]
\[= \sqrt{25} = 5;\]
\[2)\ DB =\]
\[= \sqrt{(1 - 4)^{2} + (1 + 3)^{2}} =\]
\[= \sqrt{25} = 5;\]
\[3)\ DC = \sqrt{(1 + 2)^{2} + (1 - 5)^{2}} =\]
\[= \sqrt{25} = 5;\]
\[4)\ AD = DB = DC = 5.\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{б)}\ 1)\ AD =\]
\[= \sqrt{(7 - 1)^{2} + ( - 8 - 0)^{2}} =\]
\[= \sqrt{100} = 10;\]
\[2)\ DB = \sqrt{(1 + 5)^{2} + (0 - 8)^{2}} =\]
\[= \sqrt{100} = 10;\]
\[3)\ DC = \sqrt{(1 - 9)^{2} + (0 - 6)^{2}} =\]
\[= \sqrt{100} = 10;\]
\[4)\ AD = DB = DC = 10.\]
\[Что\ и\ требовалось\ доказать.\]