\[\boxed{\mathbf{1135.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[A_{1}A_{2}A_{3}A_{4}A_{5}A_{6} - правильный\ \]
\[шестиугольник;\]
\[S_{кр} = 36\pi\ см^{2}.\]
\[\mathbf{Найти:}\]
\[A_{1}A_{2} - ?\ \]
\[S_{6} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{кр} = \pi R^{2}\]
\[36\pi = \pi R^{2}\]
\[R^{2} = \frac{36\pi}{\pi} = 36\]
\[R = \sqrt{36} = 6\ см.\]
\[2)\ a_{6} = 2R \bullet \sin\frac{180{^\circ}}{6} =\]
\[= 2 \bullet 6 \bullet \cos{30{^\circ}} = \frac{1}{2} \bullet 2 \bullet 6 = 6\ см.\]
\[3)\ P = a_{6} \bullet 6 = 6 \bullet 6 = 36\ см.\]
\[4)\ r = R \bullet \cos\frac{180{^\circ}}{6} =\]
\[= 6 \bullet \cos{30{^\circ}} = 6 \bullet \frac{\sqrt{3}}{2} = 3\sqrt{3}\ см.\]
\[5)\ S_{6} = \frac{1}{2} \bullet P \bullet r = \frac{1}{2} \bullet 36 \bullet 3\sqrt{3} =\]
\[= 18 \bullet 3\sqrt{3} = 54\sqrt{3}\ см^{2}.\]
\[Ответ:A_{1}A_{2} = 6\ см;\]
\[S_{6} = 54\sqrt{3}\ см^{2}.\]