\[\boxed{\mathbf{581.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[AC = 165\ см;\]
\[BC = 12\ см;\]
\[AD = 120\ см;\]
\[DE = 4,8\ м;\]
\[\angle 1 = \angle 2.\]
\[\mathbf{Найти:}\]
\[FE - ?\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}ABD\sim\mathrm{\Delta}DFE\ \]
\[(по\ двум\ углам):\]
\[\angle A = \angle E = 90{^\circ}\ (по\ условию);\]
\[\angle 1 = \angle 2\ (по\ условию).\]
\[Отсюда:\]
\[\frac{\text{AB}}{\text{FE}} = \frac{\text{AD}}{\text{DE}} = \frac{\text{BD}}{\text{DF}} = k.\]
\[2)\ AB = AC - BC = 165 - 12 =\]
\[= 153\ см.\]
\[3)\ \frac{153}{\text{FE}} = \frac{120}{480}\]
\[\frac{153}{\text{FE}} = \frac{1}{4}\]
\[FE = 153 \bullet 4 = 612\ см.\]
\[Ответ:FE = 612\ см.\]
\[\boxed{\mathbf{581.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ a = 5;b = 8:\]
\[c^{2} = 36 + 64 = 100\]
\[c = 10.\]
\[\textbf{б)}\ a = 5;b = 6:\]
\[c = 25 + 36 = 61\]
\[c = \sqrt{61}.\]
\[\textbf{в)}\ a = 3;b = \frac{4}{7}:\]
\[c^{2} = \frac{9}{49} + \frac{16}{49} = \frac{25}{49}\]
\[c = \frac{5}{7}.\]
\[\textbf{г)}\ a = 8;b = 8\sqrt{3}:\]
\[c^{2} = 64 + 64 \bullet 3 = 256\]
\[c = 16.\]