ГДЗ по геометрии 9 класс Мерзляк Задание 275

\[Схематический\ рисунок.\]

\[Дано:\]

\[\mathrm{\Delta}ABC - равносторонний;\]

\[O - центр\ опис.\ окружности;\]

\[AB = a.\]

\[Найти:\]

\[S_{\text{BC}}.\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[AB = BC = a,\ \ \ \angle A = \angle C = 60{^\circ};\]

\[R = \frac{\text{AB}}{2\sin{\angle C}} = \frac{a}{2\sin{60{^\circ}}} = \frac{a}{\sqrt{3}};\]

\[\angle BOC = \frac{360{^\circ}}{3} = 120{^\circ}.\]

\[2)\ В\ \mathrm{\Delta}BOC:\]

\[S_{\text{BOC}} = \frac{1}{2} \bullet \left( \frac{a}{\sqrt{3}} \right)^{2} \bullet \sin{120{^\circ}} =\]

\[= \frac{a^{2}}{6} \bullet \frac{\sqrt{3}}{2} = \frac{a^{2}\sqrt{3}}{12}.\]

\[3)\ Окружность:\]

\[S_{\text{BOC}} = \frac{\pi R^{2}\alpha}{360{^\circ}} = \frac{\pi \bullet AB^{2} \bullet 120{^\circ}}{360{^\circ}} =\]

\[= \frac{a^{2}\pi}{9};\]

\[S_{\text{BC}} = \frac{a^{2}\pi}{9} - \frac{a^{2}\sqrt{3}}{12} =\]

\[= \frac{4a^{2}\pi - 3a^{2}\sqrt{3}}{36} =\]

\[= \frac{a^{2}\left( 4\pi - 3\sqrt{3} \right)}{36}.\]

\[Ответ:\ \ \frac{a^{2}\left( 4\pi - 3\sqrt{3} \right)}{36}.\]