ГДЗ по геометрии 7 класс Атанасян Задание 1096

\[\boxed{\mathbf{1096.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\mathbf{Дано:}\]

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

\[ABCD - квадрат;\]

\[ABCDEF - правильный\ \]

\[многоугольник;\]

\[a_{3} = a_{4} = a_{6}.\]

\[\mathbf{Найти:}\]

\[S_{3}\ :S_{4}\ :S_{6}.\]

\[\mathbf{Решение.}\]

\[1)\ S = \frac{1}{2}P \bullet r.\]

\[2)\ n = 3:\]

\[a_{3} = 2R \bullet \sin{60{^\circ}}\]

\[a_{3} = 2R \bullet \frac{\sqrt{3}}{2} = R\sqrt{3}\]

\[P = 3R\sqrt{3}.\]

\[r = R \bullet \cos{60{^\circ}} = \frac{a_{3}}{\sqrt{3}} \bullet \frac{1}{2} = \frac{a_{3}}{2\sqrt{3}} =\]

\[= \frac{a_{3}\sqrt{3}}{6}.\]

\[S_{3} = \frac{1}{2} \bullet 3\sqrt{3}R \bullet \frac{a_{3}\sqrt{3}}{6} =\]

\[= \frac{1}{2} \bullet 3\sqrt{3} \bullet \frac{a_{3}}{\sqrt{3}} \bullet \frac{a_{3}\sqrt{3}}{6} =\]

\[= \frac{9\left( a_{3} \right)^{2}}{12\sqrt{3}} = \frac{3\left( a_{3} \right)^{2}}{4\sqrt{3}} = \frac{\sqrt{3}}{4}\left( a_{3} \right)^{2}.\]

\[3)\ n = 4:\]

\[a_{4} = 2R \bullet \sin{45{^\circ}}\]

\[a_{4} = 2R \bullet \frac{\sqrt{2}}{2} = R\sqrt{2}\]

\[P = 4R\sqrt{2}.\]

\[r = R \bullet \cos{45{^\circ}} = \frac{a_{4}}{\sqrt{2}} \bullet \frac{\sqrt{2}}{2} = \frac{a_{4}}{2}.\]

\[S_{4} = \frac{1}{2} \bullet 4\sqrt{2}R \bullet \frac{a_{4}}{2} =\]

\[= \frac{1}{2} \bullet 4\sqrt{2} \bullet \frac{a_{4}}{\sqrt{2}} \bullet \frac{a_{4}}{2} = \frac{4\sqrt{2}\left( a_{3} \right)^{2}}{4\sqrt{2}} =\]

\[= \left( a_{4} \right)^{2} = \left( a_{3} \right)^{2}.\]

\[4)\ n = 6:\]

\[a_{6} = 2R \bullet \sin{30{^\circ}}\]

\[a_{6} = 2R \bullet \frac{1}{2} = R\]

\[P = 6 \bullet R;\]

\[r = R \bullet \cos{30{^\circ}} = R \bullet \frac{\sqrt{3}}{2} = \frac{{\sqrt{3a}}_{6}}{2}.\]

\[S_{6} = \frac{1}{2} \bullet 6R \bullet \frac{{\sqrt{3a}}_{6}}{2} =\]

\[= \frac{1}{2} \bullet 6 \bullet a_{6} \bullet \frac{\sqrt{3}}{2} \bullet a_{6} =\]

\[= \frac{6\sqrt{3}\left( a_{6} \right)^{2}}{4} =\]

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

\[5)\ \frac{S_{3}}{S_{4}} = \frac{\sqrt{3}\left( a_{3} \right)^{2}}{4\left( a_{3} \right)^{2}} = \frac{\sqrt{3}}{4}.\]

\[6)\ \frac{S_{4}}{S_{6}} = \frac{2\left( a_{3} \right)^{2}}{3\sqrt{3}\left( a_{3} \right)^{2}} = \frac{2}{3\sqrt{3}} =\]

\[= \frac{2 \bullet 2}{6\sqrt{3}} = \frac{4}{6\sqrt{3}}.\]

\[7)\ S_{3}\ :\text{\ S}_{4}\ :\ S_{6} = \sqrt{3}\ :\ 4\ :\ 6\sqrt{3}.\]

\[Ответ:\ S_{3}\ :\text{\ S}_{4}\ :\ S_{6} =\]

\[= \sqrt{3}\ :\ 4\ :\ 6\sqrt{3}.\]