Решебник по геометрии 9 класс Мерзляк Задание 277

\[Схематический\ рисунок.\]

\[Дано:\]

\[ABCD - квадрат;\]

\[R_{окр} = AB = a.\]

\[Найти:\]

\[8S_{\text{BF}}.\]

\[Решение.\]

\[1)\ ABCD - квадрат:\]

\[AC \cap BD = F;\]

\[EF\bot AB;\ \ \ E \in AB;\]

\[AE = BE = EF = \frac{1}{2}AB = \frac{a}{2}.\]

\[2)\ В\ \mathrm{\Delta}BEF:\]

\[S_{\text{BEF}} = \frac{1}{2} \bullet BE \bullet EF = \frac{1}{2} \bullet \frac{a^{2}}{4} = \frac{a^{2}}{8}.\]

\[3)\ E - центр\ окружности\ AB:\]

\[S_{\text{BEF}} = \frac{\pi R^{2} \bullet 90{^\circ}}{360{^\circ}} = \frac{\pi \bullet a^{2}}{4 \bullet 4} = \frac{\pi a^{2}}{16};\]

\[S_{\text{BF}} = \frac{\pi a^{2}}{16} - \frac{a^{2}}{8} = \frac{a^{2}(\pi - 2)}{16};\]

\[8S_{\text{BF}} = \frac{a^{2}(\pi - 2)}{2} = a^{2}\left( \frac{\pi}{2} - 1 \right).\]

\[Ответ:\ \ a^{2}\left( \frac{\pi}{2} - 1 \right).\]