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МатематикаРешебник по геометрии 8 класс Атанасян ФГОС Задание 1097
Задание 1097
\[\boxed{\mathbf{1097.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}\ и\ \]
\[B_{1}B_{2}B_{3}B_{4}B_{5}B_{6} -\]
\[правильный\ шестиугольник.\]
\[\mathbf{Найти:}\]
\[S_{A}\ :S_{B}.\]
\[\mathbf{Решение.}\]
\[1)\ Пусть\ \widehat{r} - радиус\ \]
\[окружности;\]
\[2)\ Описанный\ шестиугольник\ \]
\[(r = \widehat{r}):\]
\[r = R \bullet \cos\frac{180{^\circ}}{6}\]
\[r = R \bullet \cos{30{^\circ}}\]
\[r = R\frac{\sqrt{3}}{2}.\]
\[R = \frac{2r}{\sqrt{3}}:\]
\[a_{6} = 2R \bullet \sin\frac{180{^\circ}}{6} =\]
\[= 2R \bullet \sin{30{^\circ}} = 2 \bullet \frac{2r}{\sqrt{3}} \bullet \frac{1}{2} = \frac{2r}{\sqrt{3}};\]
\[P = 6 \bullet \frac{2r}{\sqrt{3}} = \frac{12r}{\sqrt{3}}.\]
\[S_{A} = \frac{1}{2} \bullet P \bullet r = \frac{1}{2} \bullet \frac{12r}{\sqrt{3}} \bullet r =\]
\[= \frac{6r^{2}}{\sqrt{3}} = 2\sqrt{3}r^{2} = 2\sqrt{3}{\widehat{r}}^{2}.\]
\[3)\ Вписанный\ шестиугольник\ \]
\[(R = \widehat{r}):\]
\[a_{6} = 2R \bullet \sin\frac{180{^\circ}}{6} =\]
\[= 2\widehat{r} \bullet \sin{30{^\circ}} = 2 \bullet \widehat{r} \bullet \frac{1}{2} = \widehat{r};\]
\[r = R \bullet \cos{\frac{180{^\circ}}{6} = R \bullet \cos{30{^\circ}}} =\]
\[= \widehat{r} \bullet \frac{\sqrt{3}}{2};\]
\[P = 6 \bullet \widehat{r}.\]
\[S_{B} = \frac{1}{2} \bullet P \bullet r = \frac{1}{2} \bullet 6 \bullet \widehat{r} \bullet \widehat{r} \bullet \frac{\sqrt{3}}{2} =\]
\[= {\widehat{r}}^{2}\frac{3\sqrt{3}}{2}.\]
\[4)\ S_{A}\ :S_{B} = \frac{2\sqrt{3} \bullet {\widehat{r}}^{2} \bullet 2}{{\widehat{r}}^{2} \bullet 3\sqrt{3}} =\]
\[= \frac{4\sqrt{3} \bullet {\widehat{r}}^{2}}{3\sqrt{3} \bullet {\widehat{r}}^{2}} = \frac{4}{3}.\]
\[Ответ:\ S_{A}\ :S_{B} = 4\ :3.\]
\[\boxed{\mathbf{1097.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[точки\ A\ и\ B;\]
\[\textbf{а)}\ 2AM^{2} - BM^{2} = 2AB^{2};\]
\[\textbf{б)}\ 2AM^{2} + 2BM^{2} = 6AB^{2}.\]
\[\mathbf{Найти:}\]
\[множество\ точек\ \text{M.}\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ Введем\ систему\ \]
\[координат:\]
\[AB \in OX;A(0;0);B(a;0);\]
\[M(x;y).\]
\[2)\ AM^{2} = x^{2} + y^{2};\ \ \ AB^{2} = a^{2};\]
\[BM^{2} = (a - x)^{2} + y^{2}.\]
\[3)\ 2\left( x^{2} + y^{2} \right) - \left( (a - x)^{2} + y^{2} \right) =\]
\[= 2a^{2}\]
\[2x^{2} + 2y^{2} - a^{2} + 2ax - x^{2} - y^{2} =\]
\[= 2a^{2}\]
\[x^{2} + y^{2} + 2ax = 3a^{2}\]
\[(x + a)^{2} + y^{2} = 4a^{2}\text{.\ }\]
\[3)\ Множество\ точек\ M:\ \]
\[окружность\ с\ центром\ \]
\[в\ точке\ ( - a;0)\ и\ R = 2a.\]
\[\textbf{б)}\ 1)\ Введем\ систему\ \]
\[координат:\]
\[AB \in OX;A(0;0);B(a;0);\]
\[M(x;y).\]
\[2)\ AM^{2} = x^{2} + y^{2};\ \ \ AB^{2} = a^{2};\]
\[BM^{2} = (a - x)^{2} + y^{2}.\]
\[4x^{2} + 4y^{2} - 4ax = 4a^{2}\]
\[\left( 4x^{2} - 4ax + a^{2} \right) - a^{2} + 4y^{2} =\]
\[= 4a^{2}\]
\[(2x - a)^{2} + 4y^{2} = 5a^{2}\]
\[\left. \ 4\left( x - \frac{a}{2} \right)^{2} + 4y^{2} = 5a^{2}\text{\ \ \ \ \ } \right|:4\]
\[\ \left( x - \frac{a}{2} \right)^{2} + y^{2} = \frac{5a^{2}}{4}.\]
\[3)\ Множество\ точек\ M:\]
\[окружность\ с\ центром\ \]
\[в\ точке\ \left( \frac{a}{2};0 \right)\ и\ R = \frac{\sqrt{5}}{2}\text{a.}\]
