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МатематикаГДЗ по геометрии 9 класс Атанасян ФГОС Задание 1070
Задание 1070
\[\boxed{\mathbf{1070.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - трапеция;\]
\[AD = 16\ см;\]
\[BC = 8\ см;\]
\[CD = 4\sqrt{7}\ см;\]
\[\angle ADC = 60{^\circ};\]
\[S_{\text{ABC}C_{1}} = S_{C_{1}\text{CD}}.\]
\[\mathbf{Найти:}\]
\[\angle CC_{1} - ?\]
\[\angle S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\sin{60{^\circ}} = \frac{\text{CH}}{4\sqrt{7}} = > CH =\]
\[= 4\sqrt{7} \bullet \frac{\sqrt{3}}{2} = 2\sqrt{21}\ см.\]
\[2)\ S_{\text{ABCD}} = \frac{BC + AD}{2} \bullet CH =\]
\[= \frac{8 + 16}{2} \bullet 2\sqrt{21} = 24\sqrt{21}\ см^{2}.\]
\[3)\ S_{C_{1}\text{CD}} = S_{\text{ABC}C_{1}} = \frac{24\sqrt{21}}{2} =\]
\[= 12\sqrt{21}\ см^{2}.\]
\[4)\ S_{C_{1}\text{CD}} = \frac{1}{2}CH \bullet C_{1}D\]
\[12\sqrt{21} = \frac{1}{2} \bullet 2\sqrt{21} \bullet C_{1}D\]
\[C_{1}D = 12\ см.\]
\[5)\ По\ теореме\ косинусов:\]
\[\left( CC_{1} \right)^{2} =\]
\[= CD^{2} + C_{1}D^{2} - 2CD \bullet C_{1}D \bullet \cos{60{^\circ}};\]
\[\left( CC_{1} \right)^{2} =\]
\[= 112 + 144 - 96\sqrt{7} \bullet \frac{1}{2} =\]
\[= 256 - 48\sqrt{7};\]
\[CC_{1} = \sqrt{256 - 48\sqrt{7}} =\]
\[= \sqrt{16\left( 16 - 3\sqrt{7} \right)} =\]
\[= 4\sqrt{16 - 3\sqrt{7}}\ \approx 11,35\ см.\]
\[Ответ:\ CC_{1} = 11,35\ см;\]
\[S_{\text{ABCD}} = 24\sqrt{21}\ см^{2}.\]
\[\boxed{\mathbf{1070.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[B \in AC;\]
\[AB = BC;\]
\[AC = 2;\]
\[\textbf{а)}\ AM^{2} + BM^{2} + CM^{2} = 50;\]
\[\textbf{б)}\ AM^{2} + 2BM^{2} + 3CM^{2} = 4.\]
\[\mathbf{Найти:}\]
\[множество\ точек\ \text{M.}\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ Введем\ систему\ \]
\[координат:\]
\[A( - 1;0);C(1;0);M(x;y);\]
\[B(0;0);\]
\[\left\{ \begin{matrix} AM^{2} = (x + 1)^{2} + y^{2} \\ BM^{2} = \left( x^{2} \right) + \left( y^{2} \right)\text{\ \ \ \ } \\ CM^{2} = (x - 1)^{2} + y^{2} \\ \end{matrix} \right.\ \]
\[3x^{2} + 3y^{2} = 48\]
\[x^{2} + y^{2} = 16.\]
\[Множество\ точек\ M:\ \]
\[окружность\ с\ центром\ в\ точке\]
\[\ B(0;0)\ и\ R = 4.\]
\[\textbf{б)}\ 1)\ Введем\ систему\ \]
\[координат:\]
\[A( - 1;0);C(1;0);M(x;y);\]
\[B(0;0);\]
\[\left\{ \begin{matrix} AM^{2} = (x + 1)^{2} + y^{2} \\ BM^{2} = \left( x^{2} \right) + \left( y^{2} \right)\text{\ \ \ \ } \\ CM^{2} = (x - 1)^{2} + y^{2} \\ \end{matrix} \right.\ \]
\[6x^{2} - 4x + 6y^{2} = 0\]
\[3x^{2} - 2x + 3y^{2} = 0\]
\[3\left( x^{2} - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9} \right) + 3y^{2} = 0\]
\[3\left( x - \frac{1}{3} \right)^{2} - \frac{1}{3} + 3y^{2} = 0\]
\[\left( x - \frac{1}{3} \right)^{2} + y^{2} = \frac{1}{9}.\]
\[Множество\ точек\ M:\]
\[окружность\ с\ центром\ в\ точке\ \]
\[\left( \frac{1}{3};0 \right)\ и\ R = \frac{1}{3}.\]
