\[\boxed{\mathbf{1006.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AB = 17\ см;\]
\[AC = 28\ см;\]
\[BH = 15\ см.\]
\[\mathbf{Найти:}\]
\[CC_{1};AA_{1};BB_{1} - медианы.\]
\[\mathbf{Решение.}\]
\[1)\ Введем\ систему\ координат:\]
\[A(0;0);B(x;15);C(28;0).\]
\[H \in AC \Longrightarrow x > 0.\]
\[2)\ AB = \sqrt{x^{2} + 15^{2}} = 17\]
\[x^{2} = 289 - 225\]
\[x^{2} = 64\]
\[x = \pm 8 \Longrightarrow B( \pm 8;15).\]
\[3)\ A_{1}\left( x_{1};y_{1} \right);B_{1}\left( x_{2};y_{2} \right);\]
\[C_{1}\left( x_{3};y_{3} \right) - середины\ BC,CA\ и\ \]
\[AB;\]
\[4)\ AA_{1} =\]
\[= \sqrt{(18 - 0)^{2} + (7,5 - 0)^{2}} =\]
\[= 19,5\ см;\]
\[BB_{1} = \sqrt{(14 + 8)^{2} + (0 - 15)^{2}} =\]
\[= 3\sqrt{29}\ см;\]
\[CC_{1} = \sqrt{(4 - 28)^{2} + (7,5 - 0)^{2}} =\]
\[= \frac{3}{2}\sqrt{281}\ см.\]
\[Ответ:\ AA_{1} = 19,5\ см;\]
\(BB_{1} = 3\sqrt{29}\ см;CC_{1} = \frac{3}{2}\sqrt{281}\ см.\)
\[\boxed{\mathbf{1006.еуроки - ответы\ на\ пятёрку}}\]
\[\overrightarrow{a} = 2\overrightarrow{i} + 3\overrightarrow{j} \Longrightarrow \overrightarrow{a}\ \left\{ 2;3 \right\}\]
\[\overrightarrow{b} = - \frac{1}{2}\overrightarrow{i} + 2\overrightarrow{j} \Longrightarrow \overrightarrow{b}\ \left\{ - \frac{1}{2};2 \right\}\]
\[\overrightarrow{c} = 8\overrightarrow{i} \Longrightarrow \ \overrightarrow{c}\ \left\{ 8;0 \right\}\]
\[\overrightarrow{d} = \overrightarrow{i} - \overrightarrow{j} \Longrightarrow \overrightarrow{d}\ \left\{ 1;\ - 1 \right\}\]
\[\overrightarrow{e} = - 2\overrightarrow{j} \Longrightarrow \overrightarrow{e}\ \left\{ 0;\ - 2 \right\}\]
\[\overrightarrow{f} = - \overrightarrow{i} \Longrightarrow \overrightarrow{f}\ \{ - 1;0\}\]