\[\boxed{\mathbf{1260.ОК\ ГДЗ - домашка\ н}а\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[\mathrm{\Delta}ABC;\ \]
\[AC = 9\ см;\ \ \]
\[\text{BC} = 12\ см;\ \]
\[AM,\ BN - медианы;\ \]
\[AM\bot BN.\]
\[Найти:\]
\[\text{AB} - ?\]
\[Решение.\]
\[1)\ Выберем\ СК\ так,\ чтобы\ \]
\[A(0;0),\ C(9;0)\text{.\ }\]
\[Пусть\ B(x;y);\ y > 0.\]
\[Тогда:\ \]
\[N\left( \frac{9}{2};0 \right);\ M\left( \frac{x + 9}{2};\ \frac{y}{2} \right).\]
\[2)\ \overrightarrow{\text{AM}}\bot\overrightarrow{\text{BN}} \Longrightarrow AM \cdot BN = 0:\]
\[\left( \frac{x + 9}{2} \right)\left( \frac{9}{2} - x \right) + \frac{y}{2} \cdot ( - y) =\]
\[= 0x( - 2)\]
\[(x + 9)\left( x - \frac{9}{2} \right) + y^{2} = 0\]
\[x^{2} + \frac{9}{2}x - \frac{81}{2} + y^{2} = 0\]
\[x^{2} + 2 \cdot \frac{9}{4}x + \frac{81}{16} + y^{2} =\]
\[= 81\left( \frac{1}{2} + \frac{1}{16} \right)\]
\[\left( x + \frac{9}{4} \right)^{2} + y^{2} = \frac{729}{16}.\]
\[Уравнение\ окружности\ с\ \]
\[центром\ K\left( - \frac{9}{4};0 \right);\]
\[радиусом\ R = \sqrt{\frac{729}{16}} = \frac{27}{4}.\]
\[Кроме\ того,\ точка\ B\ лежит\ на\ \]
\[окружности\ с\ центром\ в\ \]
\[точке\ C(9;0),\ радиусом\ \ \]
\[BC = 12\ :(x - 9)^{2} + y^{2} = 144.\]
\[3)\ Получаем\ систему\ \]
\[уравнений:\]
\[\left\{ \begin{matrix} (x - 9)^{2} + y^{2} = 144 \\ \left( x + \frac{9}{4} \right)^{2} + y^{2} = \frac{729}{16} \\ \end{matrix} \right.\ \]
\[4)\ (x - 9)^{2} - \left( x + \frac{9}{4} \right)^{2} =\]
\[= 144 - \frac{729}{16};\]
\[\left( x - 9 - x - \frac{9}{4} \right)\left( x - 9 + x + \frac{9}{4} \right) =\]
\[= \frac{12^{2} \cdot 4^{2} - 27^{2}}{16}\]
\[- \frac{45}{4}\left( 2x - \frac{27}{4} \right) =\]
\[= \frac{(48 - 27)(48 + 27)}{16}\ \ | \cdot 16\]
\[- 45(8x - 27) = 21 \cdot 75\]
\[8x - 27 = - 35\]
\[8x = - 8\]
\[x = - 1.\]
\[5)\ Вторая\ координата:\]
\[y^{2} = 144 - (x - 9)^{2} =\]
\[= 144 - 100 = 44 \Longrightarrow y = 2\sqrt{11}.\]
\[B\left( - 1;2\sqrt{11} \right).\]
\[6)\ AB = \sqrt{( - 1)^{2} + \left( 2\sqrt{11} \right)^{2}} =\]
\[= \sqrt{45} = 3\sqrt{5}\ (см).\]
\[Ответ:AB = 3\sqrt{5}\ см.\]