\[\boxed{\mathbf{745.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - прямоугольник;\]
\[AB = 3\ см;\]
\[BC = 4\ см;\]
\[M - середина\ \text{AB.}\]
\[\mathbf{Найти:}\]
\[\left| \overrightarrow{\text{AB}} \right|;\left| \overrightarrow{\text{BC}} \right|;\]
\[\left| \overrightarrow{\text{DC}} \right|;\left| \overrightarrow{\text{MC}} \right|;\]
\[\left| \overrightarrow{\text{MA}} \right|;\left| \overrightarrow{\text{CB}} \right|;\]
\[\left| \overrightarrow{\text{AC}} \right| - ?\]
\[\mathbf{Решение.}\]
\[1)\ \left| \overrightarrow{\text{AB}} \right| = 3\ см.\]
\[2)\ \left| \overrightarrow{\text{BC}} \right| = \left| \overrightarrow{\text{CB}} \right| = 4\ см.\]
\[4)\ \left| \overrightarrow{\text{MA}} \right| = \frac{1}{2}\left| \overrightarrow{\text{BA}} \right| = \frac{3}{2} = 1,5\ см.\]
\[5)\ По\ теореме\ Пифагора:\]
\[MC = \sqrt{CB^{2} + BM^{2}} =\]
\[= \sqrt{16 + 2,25} = \sqrt{18,25} =\]
\[= \sqrt{1825 \bullet 0,01} =\]
\[= \sqrt{25 \bullet 73 \bullet 0,01} = 5 \bullet 0,1\sqrt{73} =\]
\[= \frac{1}{2}\sqrt{73}\ см.\]
\[\left| \overrightarrow{\text{MC}} \right| = \frac{1}{2}\sqrt{73}\ см.\]
\[6)\ AC = \sqrt{BC^{2} + AB^{2}} =\]
\[= \sqrt{16 + 9} = \sqrt{25} = 5\ см;\]
\[\left| \overrightarrow{\text{AC}} \right| = 5\ см.\]