Решебник по геометрии 9 класс Мерзляк Задание 259

\[R = \ 5\ см.\]

\[1)\ a = 45{^\circ}:\]

\[S_{с} = \frac{a\pi R^{2}}{360{^\circ}} - \frac{1}{2}R^{2} \bullet \sin{45{^\circ}} =\]

\[= \frac{45{^\circ} \bullet \pi \bullet 25}{360{^\circ}} - \frac{25}{2} \bullet \frac{\sqrt{2}}{2} =\]

\[= \frac{25\pi - 25 \bullet 2\sqrt{2}}{8} = \frac{25\left( \pi - 2\sqrt{2} \right)}{8}.\]

\[Ответ:\ \ \frac{25\left( \pi - 2\sqrt{2} \right)}{8}\ см^{2}.\]

\[2)\ a = 330{^\circ}:\]

\[S_{с} = \frac{a\pi R^{2}}{360{^\circ}} - \frac{1}{2}R^{2} \bullet \sin{330{^\circ}} =\]

\[= \frac{330{^\circ} \bullet \pi \bullet 25}{360{^\circ}} + \frac{25}{2} \bullet \frac{1}{2} =\]

\[= \frac{275\pi + 25 \bullet 3}{12} = \frac{25(11\pi + 3)}{12}.\]

\[Ответ:\ \ \frac{25(11\pi + 3)}{12}\ см^{2}.\]