ГДЗ по геометрии 8 класс Атанасян Задание 1097

\[\boxed{\mathbf{1097.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}\ и\ \]

\[B_{1}B_{2}B_{3}B_{4}B_{5}B_{6} -\]

\[правильный\ шестиугольник.\]

\[\mathbf{Найти:}\]

\[S_{A}\ :S_{B}.\]

\[\mathbf{Решение.}\]

\[1)\ Пусть\ \widehat{r} - радиус\ \]

\[окружности;\]

\[2)\ Описанный\ шестиугольник\ \]

\[(r = \widehat{r}):\]

\[r = R \bullet \cos\frac{180{^\circ}}{6}\]

\[r = R \bullet \cos{30{^\circ}}\]

\[r = R\frac{\sqrt{3}}{2}.\]

\[R = \frac{2r}{\sqrt{3}}:\]

\[a_{6} = 2R \bullet \sin\frac{180{^\circ}}{6} =\]

\[= 2R \bullet \sin{30{^\circ}} = 2 \bullet \frac{2r}{\sqrt{3}} \bullet \frac{1}{2} = \frac{2r}{\sqrt{3}};\]

\[P = 6 \bullet \frac{2r}{\sqrt{3}} = \frac{12r}{\sqrt{3}}.\]

\[S_{A} = \frac{1}{2} \bullet P \bullet r = \frac{1}{2} \bullet \frac{12r}{\sqrt{3}} \bullet r =\]

\[= \frac{6r^{2}}{\sqrt{3}} = 2\sqrt{3}r^{2} = 2\sqrt{3}{\widehat{r}}^{2}.\]

\[3)\ Вписанный\ шестиугольник\ \]

\[(R = \widehat{r}):\]

\[a_{6} = 2R \bullet \sin\frac{180{^\circ}}{6} =\]

\[= 2\widehat{r} \bullet \sin{30{^\circ}} = 2 \bullet \widehat{r} \bullet \frac{1}{2} = \widehat{r};\]

\[r = R \bullet \cos{\frac{180{^\circ}}{6} = R \bullet \cos{30{^\circ}}} =\]

\[= \widehat{r} \bullet \frac{\sqrt{3}}{2};\]

\[P = 6 \bullet \widehat{r}.\]

\[S_{B} = \frac{1}{2} \bullet P \bullet r = \frac{1}{2} \bullet 6 \bullet \widehat{r} \bullet \widehat{r} \bullet \frac{\sqrt{3}}{2} =\]

\[= {\widehat{r}}^{2}\frac{3\sqrt{3}}{2}.\]

\[4)\ S_{A}\ :S_{B} = \frac{2\sqrt{3} \bullet {\widehat{r}}^{2} \bullet 2}{{\widehat{r}}^{2} \bullet 3\sqrt{3}} =\]

\[= \frac{4\sqrt{3} \bullet {\widehat{r}}^{2}}{3\sqrt{3} \bullet {\widehat{r}}^{2}} = \frac{4}{3}.\]

\[Ответ:\ S_{A}\ :S_{B} = 4\ :3.\]