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МатематикаРешебник по геометрии 11 класс. Атанасян ФГОС 755
755
\[\boxed{\mathbf{755.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[ABCD - ромб;\]
\[DB\bot AC;\]
\[N - середина\ BC;\]
\[AC = 8;\]
\[DB = MO = 6.\]
\[Решение.\]
\[По\ теореме\ Пифагора\ и\ \]
\[свойству\ диагоналей\ ромба:\]
\[BC = \sqrt{AO^{2} + OB^{2}} =\]
\[= \sqrt{\left( \frac{\text{AC}}{2} \right)^{2} + \left( \frac{\text{BD}}{2} \right)^{2}} = 5.\]
\[\cos{\angle OBC} = \frac{\text{OB}}{\text{BC}} = \frac{3}{5};\]
\[\cos{\angle OCB} = \frac{\text{OC}}{\text{BC}} = \frac{4}{5}.\]
\[BN = CN = 2,5.\]
\[\textbf{а)}\ \overrightarrow{\text{MN}} = \overrightarrow{\text{MO}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{BN}}\]
\[\overrightarrow{\text{BC}} \cdot \overrightarrow{\text{MN}} =\]
\[= \overrightarrow{\text{BC}}\left( \overrightarrow{\text{MO}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{BN}} \right) =\]
\[= \overrightarrow{\text{BC}} \cdot \overrightarrow{\text{MO}} + \overrightarrow{\text{BC}} \cdot \overrightarrow{\text{OB}} + \overrightarrow{\text{BC}} \cdot \overrightarrow{\text{BN}} =\]
\[= - 9 + 12,5 = 3,5.\]
\[\overrightarrow{\text{BC}} \cdot \overrightarrow{\text{MN}} =\]
\[= \left| \overrightarrow{\text{BC}} \right| \cdot \left| \overrightarrow{\text{MN}} \right| \cdot \cos{\angle\left( \overrightarrow{\text{BC}};\overrightarrow{\text{MN}} \right)}\]
\[MN = \sqrt{MO^{2} + ON^{2}} =\]
\[= \sqrt{36 + 6,25} = 6,5;\]
\[\cos{\angle\left( \overrightarrow{\text{BC}};\overrightarrow{\text{MN}} \right)} = \frac{3,5}{5 \cdot 6,5} = \frac{7}{65}.\]
\[\textbf{б)}\ \left| \overrightarrow{\text{DC}} \right| = \left| \overrightarrow{\text{BC}} \right| = 5;\]
\[\overrightarrow{\text{MN}} = \overrightarrow{\text{MO}} + \overrightarrow{\text{OC}} + \overrightarrow{\text{CN}};\]
\[\overrightarrow{\text{DC}} \cdot \overrightarrow{\text{MN}} =\]
\[= \overrightarrow{\text{DC}}\left( \overrightarrow{\text{MO}} + \overrightarrow{\text{OC}} + \overrightarrow{\text{CN}} \right) =\]
\[\overrightarrow{\text{DC}} \cdot \overrightarrow{\text{MO}} + \overrightarrow{\text{DC}} \cdot \overrightarrow{\text{OC}} + \overrightarrow{\text{DC}} \cdot \overrightarrow{\text{CN}} =\]
\[= 16 - 3,5 = 12,5.\]
\[\cos{\angle\left( \overrightarrow{\text{DC}};\overrightarrow{\text{MN}} \right)} = \frac{12,5}{5 \cdot 6,5} = \frac{5}{13}.\]
\[\textbf{в)}\ \overrightarrow{\text{AC}} \cdot \overrightarrow{\text{MN}} =\]
\[= \left( \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} \right)\left( \overrightarrow{\text{MO}} + \overrightarrow{\text{BO}} + \overrightarrow{\text{BN}} \right) =\]
\[= 0 + 9 + 3,5 - 9 + 0 + 12,5 =\]
\[= 16;\]
\[\cos{\angle\left( \overrightarrow{\text{AC}};\overrightarrow{\text{MN}} \right)} = \frac{16}{8 \cdot 6,5} = \frac{4}{13}.\]
\[\textbf{г)}\ \overrightarrow{\text{DB}} \cdot \overrightarrow{\text{MN}} =\]
\[= \left( \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} \right)\left( \overrightarrow{\text{MO}} + \overrightarrow{\text{OC}} + \overrightarrow{\text{CN}} \right) =\]
\[= 16 - 16 - 3,5 + 12,5 = 9.\]
\[\cos{\angle\left( \overrightarrow{\text{DB}};\overrightarrow{\text{MN}} \right)} = \frac{9}{39} = \frac{3}{13}\text{.\ }\]
