Решебник по геометрии 11 класс. Атанасян ФГОС 673

\[\boxed{\mathbf{673.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[ABCD - трапеция.\]

\[A(4;4;0);\ \ B(0;0;0);\ \ C(0;3;4);\ \ \]

\[D(1;4;4).\]

\[\overrightarrow{\text{AB}}\left\{ - 4;4;0 \right\}:\]

\[\left| \overrightarrow{\text{AB}} \right| = \sqrt{4^{2} + 4^{2} + 0^{2}} = \sqrt{32} =\]

\[= 4\sqrt{2}.\]

\[\overrightarrow{\text{BC}}\left\{ 0;3;4 \right\}:\]

\[\left| \overrightarrow{\text{BC}} \right| = \sqrt{0^{2} + 3^{2} + 4^{2}} = \sqrt{25} =\]

\[= 5.\]

\[\overrightarrow{\text{CD}}\left\{ 1;1;0 \right\}:\]

\[\left| \overrightarrow{\text{CD}} \right| = \sqrt{1^{2} + 1^{2} + 0^{2}} = \sqrt{2}.\]

\[\overrightarrow{\text{DA}}\left\{ 3;0;4 \right\}:\]

\[\left| \overrightarrow{\text{DA}} \right| = \sqrt{3^{2} + 0^{2} + 4^{2}} = \sqrt{25} =\]

\[= 5.\]

\[\left| \overrightarrow{\text{BC}} \right| = \left| \overrightarrow{\text{DA}} \right| = 5;\ \ \ \overrightarrow{\text{AB}}\ и\ \overrightarrow{\text{CD}} -\]

\[коллинеарны:\]

\[ABCD - равнобедренная\ \]

\[трапеция.\]

\[Что\ и\ требовалось\ доказать.\]