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МатематикаГДЗ по геометрии 7 класс Атанасян ФГОС Задание 1096
Задание 1096
\[\boxed{\mathbf{1096.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - равносторонний;\]
\[ABCD - квадрат;\]
\[ABCDEF - правильный\ \]
\[многоугольник;\]
\[a_{3} = a_{4} = a_{6}.\]
\[\mathbf{Найти:}\]
\[S_{3}\ :S_{4}\ :S_{6}.\]
\[\mathbf{Решение.}\]
\[1)\ S = \frac{1}{2}P \bullet r.\]
\[2)\ n = 3:\]
\[a_{3} = 2R \bullet \sin{60{^\circ}}\]
\[a_{3} = 2R \bullet \frac{\sqrt{3}}{2} = R\sqrt{3}\]
\[P = 3R\sqrt{3}.\]
\[r = R \bullet \cos{60{^\circ}} = \frac{a_{3}}{\sqrt{3}} \bullet \frac{1}{2} = \frac{a_{3}}{2\sqrt{3}} =\]
\[= \frac{a_{3}\sqrt{3}}{6}.\]
\[S_{3} = \frac{1}{2} \bullet 3\sqrt{3}R \bullet \frac{a_{3}\sqrt{3}}{6} =\]
\[= \frac{1}{2} \bullet 3\sqrt{3} \bullet \frac{a_{3}}{\sqrt{3}} \bullet \frac{a_{3}\sqrt{3}}{6} =\]
\[= \frac{9\left( a_{3} \right)^{2}}{12\sqrt{3}} = \frac{3\left( a_{3} \right)^{2}}{4\sqrt{3}} = \frac{\sqrt{3}}{4}\left( a_{3} \right)^{2}.\]
\[3)\ n = 4:\]
\[a_{4} = 2R \bullet \sin{45{^\circ}}\]
\[a_{4} = 2R \bullet \frac{\sqrt{2}}{2} = R\sqrt{2}\]
\[P = 4R\sqrt{2}.\]
\[r = R \bullet \cos{45{^\circ}} = \frac{a_{4}}{\sqrt{2}} \bullet \frac{\sqrt{2}}{2} = \frac{a_{4}}{2}.\]
\[S_{4} = \frac{1}{2} \bullet 4\sqrt{2}R \bullet \frac{a_{4}}{2} =\]
\[= \frac{1}{2} \bullet 4\sqrt{2} \bullet \frac{a_{4}}{\sqrt{2}} \bullet \frac{a_{4}}{2} = \frac{4\sqrt{2}\left( a_{3} \right)^{2}}{4\sqrt{2}} =\]
\[= \left( a_{4} \right)^{2} = \left( a_{3} \right)^{2}.\]
\[4)\ n = 6:\]
\[a_{6} = 2R \bullet \sin{30{^\circ}}\]
\[a_{6} = 2R \bullet \frac{1}{2} = R\]
\[P = 6 \bullet R;\]
\[r = R \bullet \cos{30{^\circ}} = R \bullet \frac{\sqrt{3}}{2} = \frac{{\sqrt{3a}}_{6}}{2}.\]
\[S_{6} = \frac{1}{2} \bullet 6R \bullet \frac{{\sqrt{3a}}_{6}}{2} =\]
\[= \frac{1}{2} \bullet 6 \bullet a_{6} \bullet \frac{\sqrt{3}}{2} \bullet a_{6} =\]
\[= \frac{6\sqrt{3}\left( a_{6} \right)^{2}}{4} =\]
\[= {\frac{3\sqrt{3}}{2}\left( a_{6} \right)^{2} = \frac{3\sqrt{3}}{2}\left( a_{3} \right)}^{2}.\]
\[5)\ \frac{S_{3}}{S_{4}} = \frac{\sqrt{3}\left( a_{3} \right)^{2}}{4\left( a_{3} \right)^{2}} = \frac{\sqrt{3}}{4}.\]
\[6)\ \frac{S_{4}}{S_{6}} = \frac{2\left( a_{3} \right)^{2}}{3\sqrt{3}\left( a_{3} \right)^{2}} = \frac{2}{3\sqrt{3}} =\]
\[= \frac{2 \bullet 2}{6\sqrt{3}} = \frac{4}{6\sqrt{3}}.\]
\[7)\ S_{3}\ :\text{\ S}_{4}\ :\ S_{6} = \sqrt{3}\ :\ 4\ :\ 6\sqrt{3}.\]
\[Ответ:\ S_{3}\ :\text{\ S}_{4}\ :\ S_{6} =\]
\[= \sqrt{3}\ :\ 4\ :\ 6\sqrt{3}.\]
\[\boxed{\mathbf{1096.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AA_{1} - медиана.\]
\[\mathbf{Доказать:}\]
\[AA_{1} = \frac{1}{2} \bullet \sqrt{2AC^{2} + 2AB^{2} - BC^{2}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ Дополнительное\ \]
\[построение:\ \]
\[продлим\ AA_{1}\ :AA_{1} = AA_{2} \Longrightarrow\]
\[\Longrightarrow \text{CAB}A_{2} - параллелограмм.\]
\[2)\ По\ свойству\ \]
\[параллелограмма:\]
\[\left( AA_{2} \right)^{2} + CB^{2} =\]
\[= AC^{2} + AB^{2} + \left( BA_{2} \right)^{2} + \left( CA_{2} \right)^{2}\]
\[\left( AA_{2} \right)^{2} = 2AC^{2} + 2AB^{2} - CB^{2}\]
\[\left( AA_{2} \right)^{2} =\]
\[= \frac{1}{4}\left( 2AC^{2} + 2AB^{2} - BC^{2} \right)\]
\[AA_{2} = \frac{1}{2}\sqrt{2AC^{2} + 2AB^{2} - BC^{2}}.\]
\[Что\ и\ требовалось\ доказать.\]
\[2)\ Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AN = CM.\]
\[\mathbf{Доказать:}\]
\[AB = BC.\]
\[\mathbf{Доказательство.}\]
\[1)\ CM =\]
\[= \frac{1}{2}\sqrt{2BC^{2} + 2AC^{2} - AB^{2}};\]
\[AN = \frac{1}{2}\sqrt{2AB^{2} + 2AC^{2} - BC^{2}};\]
\[2)\ AN = CM\ (по\ условию):\]
\[\frac{1}{2}\sqrt{2BC^{2} + 2AC^{2} - AB^{2}} =\]
\[= \frac{1}{2}\sqrt{2AB^{2} + 2AC^{2} - BC^{2}}\]
\[2BC^{2} + 2AC^{2} - AB^{2} =\]
\[= 2AB^{2} + 2AC^{2} - BC^{2}\]
\[3BC^{2} = 3AB^{2} \Longrightarrow BC = AD:\]
\[\mathrm{\Delta}ABC - равнобедренный\]
\[Что\ и\ требовалось\ доказать.\]
