Решебник по геометрии 10 класс Атанасян ФГОС 716

\[\boxed{\mathbf{716.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\angle BAC = 90{^\circ};\]

\[\angle DAB = 60{^\circ};\]

\[\angle DAC = 45{^\circ};\]

\[DA = 5\ см;\ \]

\[AB = 4\ см;\]

\[AC = 3\ см;\]

\[M - точка\ пересечения\ медиан\ \]

\[⊿DBC.\]

\[Найти:\]

\[\text{AM.}\]

\[Решение.\]

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

\[N\left( 2;\frac{3}{2};0 \right);\]

\[D\left( 5\cos{60{^\circ}};5\cos{45{^\circ}};5\sin\gamma \right);\]

\[\overrightarrow{\text{AD}}\left\{ \frac{5}{2};\frac{5\sqrt{2}}{2};\sin\gamma \right\}.\]

\[\overrightarrow{\text{AD}} \cdot \overrightarrow{\text{AD}} = 25:\]

\[\left( \frac{5}{2} \right)^{2} + \left( \frac{5\sqrt{2}}{2} \right)^{2} + \left( 5\sin\gamma \right)^{2}\]

\[\sin\gamma = \sqrt{1 - \frac{1}{4} - \frac{2}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2};\]

\[\gamma = 30{^\circ}.\]

\[\overrightarrow{\text{AD}}\left( \frac{5}{2};\frac{5\sqrt{2}}{2};\frac{5}{2} \right);\ \ \angle DAN = 30{^\circ};\ \ \]

\[\overrightarrow{\text{AN}}\left\{ 2;\frac{3}{2};0 \right\}.\]

\[\overrightarrow{\text{AN}} = \sqrt{4 + \frac{9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2};\]

\[\overrightarrow{\text{AM}} = \overrightarrow{\text{AN}} + \overrightarrow{\text{NM}} = \overrightarrow{\text{AN}} + \frac{1}{3}\overrightarrow{\text{ND}} =\]

\[= \overrightarrow{\text{AN}} + \frac{1}{3}\left( \overrightarrow{\text{AD}} - \overrightarrow{\text{AN}} \right) =\]

\[= \frac{1}{3}\overrightarrow{\text{AD}} + \frac{2}{3}\overrightarrow{\text{AN}};\]

\[\overrightarrow{\text{AM}}\left\{ \frac{13}{6};\frac{5\sqrt{2} + 6}{6};\frac{5}{6} \right\}:\]

\[\left| \overrightarrow{\text{AM}} \right| =\]

\[= \sqrt{\frac{169}{36} + \frac{\left( 5\sqrt{2} + 6 \right)^{2}}{36} + \frac{25}{36}} =\]

\[= \sqrt{\frac{169 + 50 + 60\sqrt{2} + 36 + 25}{36}} =\]

\[= \frac{\sqrt{280 + 60\sqrt{2}}}{6} =\]

\[= \frac{2}{6}\sqrt{70 + 15\sqrt{2}} =\]

\[= \frac{1}{3}\sqrt{70 + 15\sqrt{2}}.\]

\[Ответ:\ \ \frac{1}{3}\sqrt{70 + 15\sqrt{2}}.\]