ГДЗ по геометрии 7 класс Атанасян Задание 938

\[\boxed{\mathbf{938.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\textbf{а)}\ \overrightarrow{a}\ \left\{ 5;9 \right\}:\ \ \ \]

\[\left| \overrightarrow{a} \right| = \sqrt{5^{2} + 9^{2}} = \sqrt{25 + 81} =\]

\[= \sqrt{106}.\]

\[\textbf{б)}\ \overrightarrow{b}\ \left\{ - 3;4 \right\}:\ \ \ \]

\[\left| \overrightarrow{b} \right| = \sqrt{( - 3)^{2} + 4^{2}} = \sqrt{9 + 16} =\]

\[= \sqrt{25} = 5.\]

\[\textbf{в)}\ \overrightarrow{c}\ \left\{ - 10; - 10 \right\}:\ \ \ \]

\[\left| \overrightarrow{c} \right| = \sqrt{( - 10)^{2} + ( - 10)^{2}} =\]

\[= \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2}.\]

\[\textbf{г)}\ \overrightarrow{d}\ \left\{ 10;17 \right\}:\ \ \ \]

\[\left| \overrightarrow{d} \right| = \sqrt{10^{2} + 17^{2}} =\]

\[= \sqrt{100 + 289} = \sqrt{389}.\]

\[\textbf{д)}\ \overrightarrow{e}\ \left\{ 11; - 11 \right\}:\ \ \ \]

\[\left| \overrightarrow{e} \right| = \sqrt{11^{2} + ( - 11)^{2}} =\]

\[= \sqrt{121 + 121} = 11\sqrt{2}.\]

\[\textbf{е)}\ \overrightarrow{f}\ \left\{ 10;0 \right\}:\ \ \ \]

\(\left| \overrightarrow{f} \right| = \sqrt{10^{2} + 0^{2}} = \sqrt{100} = 10.\)