\[\boxed{\mathbf{693}\mathbf{.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - прямоугольный;\]
\[прямоугольн;\]
\[\angle C = 90{^\circ};\]
\[\textbf{а)}\ AB = 26\ см;\]
\[r = 4\ см;\]
\[\textbf{б)}\ AD = 5\ см;\]
\[DB = 12\ см.\]
\[\mathbf{Найти:}\]
\[P_{\text{ABC}} - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ FOEC - четырехугольник;\ \]
\[\angle OFC = \angle FOE = \angle OEC =\]
\[= \angle ECF = 90{^\circ};\]
\[\ FO = OE.\]
\[Значит:\ \]
\[FOEC - квадрат \Longrightarrow\]
\[\Longrightarrow FC = CE = OE = FO = 4\ см.\]
\[2)\ AF = AD;\ \ BD = BE\ \]
\[(по\ теореме\ о\ касательных).\]
\[3)\ AB = AD + DB.\]
\[4)\ P_{\text{ABC}} = AB + BC + AC =\]
\[= AB + CE + EB + AF + FC;\]
\[P_{\text{ABC}} =\]
\[= AB + CE + FC + BD + AD =\]
\[= 26 + 4 + 4 + 4 + AB =\]
\[= 26 + 4 + 4 + 26 = 60\ см.\]
\[\textbf{б)}\ 1)\ AB = AD + DB = 5 + 12 =\]
\[= 17\ см.\]
\[2)\ По\ теореме\ о\ касательных\ к\ \]
\[окружности:\]
\[AF = AD = 5\ см;\ \]
\[BD = BE = 12\ см.\]
\[3)\ AC = AF + r = 5 + r;\ \ \ \]
\[BC = BE + r = 12 + r.\]
\[4)\ По\ теореме\ Пифагора:\]
\[(5 + r)^{2} + (12 + r)^{2} = 17^{2}\]
\[25 + 10r + r^{2} + 144 + 24r + r^{2} =\]
\[= 289\]
\[2r^{2} + 34r - 120 = 0\]
\[r^{2} + 17r - 60 = 0\]
\[По\ теореме\ Виета:\]
\[r_{1} + r_{2} = - 17;\ \ \ r_{1} \bullet r_{2} = - 60\ \]
\[r_{1} = 3;\ \ r_{2} = - 20.\ \]
\[Значит:\ \]
\[r = 3\ см.\]
\[5)\ AC = 5 + 3 = 8\ см;\]
\[BC = 12 + 3 = 15\ см.\]
\[5)\ P_{\text{ABC}} = AB + BC + AC =\]
\[= 17 + 15 + 8 = 40\ см.\]
\[Ответ:а)\ 60\ см;\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ 40\ см.\ \]