\[Схематический\ рисунок.\]
\[Дано:\]
\[O - центр\ окружности;\]
\[BH - высота;\]
\[AB = BC;\]
\[BH = 27\ см;\]
\[\angle A = 70{^\circ}.\]
\[Найти:\]
\[\cup EF.\]
\[Решение:\]
\[1)\ В\ \mathrm{\Delta}ABC:\]
\[AB = BC;\ \ \]
\[\angle A = \angle C = 70{^\circ};\]
\[\angle B = 180{^\circ} - \angle A - \angle C = 40{^\circ}.\]
\[2)\ Окружность:\]
\[R = \frac{1}{2}d = \frac{1}{2}BH = \frac{27}{2};\]
\[\angle EOF = 2\angle EBF = 80{^\circ};\]
\[C = 2\pi R = 2\pi \bullet \frac{27}{2} = 27\pi;\]
\[\cup EF = 27\pi \bullet \frac{80{^\circ}}{360{^\circ}} = 6\pi.\]
\[Ответ:\ \ 6\pi\ см.\]