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МатематикаГДЗ по геометрии 7 класс Атанасян ФГОС Задание 1077
Задание 1077
\[\boxed{\mathbf{1077.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисункт\ по\ условию\ задачи:\]
\[\mathbf{а)\ }\]
\[\mathbf{б)\ }\]
\[\mathbf{Дано:\ }\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}A_{1}B_{1}C_{1}:\]
\[AB = c;BC = a;CA = b;\]
\[A_{1}B_{1} = c_{1};B_{1}C_{1} = a_{1};C_{1}A_{1} = b_{1}.\]
\[k - коэффициент\ подобия:\]
\[k = \frac{a}{a_{1}} = \frac{b}{b_{1}} = \frac{c}{c_{1}}.\]
\[Доказать:\]
\[\textbf{а)}\ k = \frac{R}{R_{1}};\]
\[\textbf{б)}\ k = \frac{r}{r_{1}}.\]
\[Доказательство.\]
\[\textbf{а)}\ 1)\frac{c}{\sin{\angle C}} = 2R;\ \ \ \ \]
\[\frac{c_{1}}{\sin{\angle C_{1}}} = 2R_{1}.\]
\[2)\ \angle C = \angle C_{1} \Longrightarrow \sin{\angle C} = \sin{\angle C_{1}}.\]
\[3)\ \frac{c_{1}}{c} = k.\]
\[4)\frac{c}{2R} = \frac{c_{1}}{2R_{1}}\]
\[\frac{c}{c_{1}} = \frac{2R}{2R_{1}} = k\]
\[\frac{R}{R_{1}} = k.\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{б)}\ 1)\ \mathrm{\Delta}ABC\sim\mathrm{\Delta}A_{1}B_{1}C_{1}\ \]
\[(по\ условию);\]
\[\mathrm{\Delta}ABO\sim\mathrm{\Delta}A_{1}B_{1}O_{1}\ \]
\[(по\ первому\ признаку);\]
\[\frac{S_{\text{ABO}}}{S_{A_{1}B_{1}O_{1}}} = \left( \frac{c}{c_{1}} \right)^{2} = k^{2}\]
\[\frac{c}{c_{1}} = k.\]
\[2)\ S_{\text{ABO}} = \frac{1}{2}c \bullet r\ и\ \]
\[S_{A_{1}B_{1}O_{1}} = \frac{1}{2}c_{1} \bullet r_{1}:\]
\[\frac{\frac{1}{2}c \bullet r}{\frac{1}{2}c_{1} \bullet r_{1}} = k^{2} \Longrightarrow \frac{r}{r_{1}} = k.\]
\[Что\ и\ требовалось\ доказать.\]
\[\mathbf{Глава\ 12.\ Длина\ окружности\ и\ площадь\ круга}\]
\[\mathbf{Параграф\ }1\mathbf{.\ Правильные\ многоугольники}\]
\[\boxed{\mathbf{1077.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ \overrightarrow{p} = 7\overrightarrow{a} - 3\overrightarrow{b};\ \ \overrightarrow{a}\left( 1; - 1 \right\};\ \]
\[\overrightarrow{b}\left\{ 5; - 2 \right\}:\]
\[\overrightarrow{p} = 7\left\{ 1; - 1 \right\} - 3\left\{ 5; - 2 \right\} =\]
\[= \left\{ 7 - 15; - 5 + 6 \right\} = \left\{ - 8; - 1 \right\};\]
\[\left| \overrightarrow{p} \right| = \sqrt{64 + 1} = \sqrt{65}.\]
\[\textbf{б)}\ \overrightarrow{p} = 4\overrightarrow{a} - 2\overrightarrow{b};\ \ \overrightarrow{a}\left( 6;3 \right\};\ \overrightarrow{b}\left\{ 5;4 \right\}:\]
\[\overrightarrow{p} = 4\left\{ 6;3 \right\} - 2\left\{ 5;4 \right\} =\]
\[= \left\{ 24 - 10;12 - 8 \right\} = \left\{ 14;4 \right\};\]
\[\left| \overrightarrow{p} \right| = \sqrt{196 + 16} = \sqrt{212} =\]
\[= 2\sqrt{53}.\]
\[\textbf{в)}\ \overrightarrow{p} = 5\overrightarrow{a} - 4\overrightarrow{b};\ \ \overrightarrow{a}\left( \frac{3}{5};\frac{1}{5} \right\};\ \]
\[\overrightarrow{b}\left\{ 6; - 1 \right\}:\]
\[\overrightarrow{p} = 5\left\{ \frac{3}{5};\frac{1}{5} \right\} - 4\left\{ 6; - 1 \right\} =\]
\[= \left\{ 3 - 24;1 + 4 \right\} = \left\{ - 21;5 \right\};\]
\[\left| \overrightarrow{p} \right| = \sqrt{441 + 25} = \sqrt{466}.\]
\[\textbf{г)}\ \overrightarrow{p} = 3\left( - 2\overrightarrow{a} - 4\overrightarrow{b} \right);\ \ \overrightarrow{a}\left( 1;5 \right\};\ \]
\[\overrightarrow{b}\left\{ - 1; - 1 \right\}:\]
\[\overrightarrow{p} = 3\left( - 2\left\{ 1;5 \right\} - 4\left\{ - 1; - 1 \right\} \right) =\]
\[= 3\left\{ - 2 + 4; - 10 + 4 \right\} =\]
\[= 3\left\{ 2; - 6 \right\} = \{ 6; - 18\};\]
\[\left| \overrightarrow{p} \right| = \sqrt{36 + 324} = \sqrt{360} =\]
\[= 6\sqrt{10}.\]
