ГДЗ по геометрии 11 класс. Атанасян ФГОС 557

\[\boxed{\mathbf{557.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[ABCD - тетраэдр;\]

\[AM = MC;\ \]

\[BN = NC;\]

\[CK = KD;\]

\[AB = 3\ см;\]

\[BC = 4\ см;\]

\[BD = 5\ см.\]

\[Найти:\]

\[длины\ векторов.\]

\[Решение.\]

\[\textbf{а)}\ \overrightarrow{\text{AB}},\ \overrightarrow{\text{BC}},\ \overrightarrow{\text{BD}},\ \overrightarrow{\text{NM}},\ \overrightarrow{\text{BN}},\ \overrightarrow{\text{NK}}:\]

\[\left| \overrightarrow{\text{AB}} \right| = AB = 3\ см;\ \ \]

\[\left| \overrightarrow{\text{BC}} \right| = BC = 4\ см;\ \]

\[\left| \overrightarrow{\text{BD}} \right| = BD = 5\ см\text{.\ \ }\]

\[N - середина\ \text{BC\ }и\ \ M -\]

\[середина\ AC:\]

\[MN - средняя\ линия\ \mathrm{\Delta}ABC.\]

\[\left| \overrightarrow{\text{NM}} \right| = NM = \frac{1}{2}AB = 1,5\ см;\]

\[\left| \overrightarrow{\text{BN}} \right| = BN = \frac{1}{2}BC = 2\ см.\]

\[N - середина\ \text{BC\ }и\ \ K -\]

\[середина\ CD:\]

\[NK - средняя\ линия\ \mathrm{\Delta}BCD.\ \]

\[\left| \overrightarrow{\text{NK}} \right| = NK = \frac{1}{2}BD = 2,5\ см.\]

\[\textbf{б)}\ \overrightarrow{\text{CB}},\ \overrightarrow{\text{BA}},\ \overrightarrow{\text{DB}},\ \overrightarrow{\text{NC}},\ \overrightarrow{\text{KN}}:\]

\[\left| \overrightarrow{\text{CB}} \right| = CB = 4\ см;\ \ \]

\[\left| \overrightarrow{\text{BA}} \right| = BA = 3\ см;\ \]

\[\left| \overrightarrow{\text{DB}} \right| = DB = 5\ см;\ \ \]

\[\left| \overrightarrow{\text{NC}} \right| = NC = \frac{1}{2}BC = 2\ см.\]

\[K - середина\ \text{CD\ }и\ \ N -\]

\[середина\ BC:\]

\[KN - средняя\ линия\ \mathrm{\Delta}BCD.\ \]

\[\left| \overrightarrow{\text{KN}} \right| = KN = \frac{1}{2}BD = 2,5\ см.\]

\[\mathbf{Ответ}\mathbf{:\ \ }а)\ 3\ см,\ 4\ см,\ 5\ см,\ 1,5\ см,\ \]

\[2\ см,\ 2,5\ см;\ \ \]

\[\textbf{б)}\ 4\ см,\ 3\ см,\ 5\ см,\ 2\ см,\ 2,5\ см.\]