Решебник по геометрии 9 класс Мерзляк Задание 273

\[Схематический\ рисунок.\]

\[Дано:\]

\[ABCD - квадрат;\]

\[O - центр\ опис.\ окружности;\]

\[AB = a.\]

\[Найти:\]

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

\[Решение.\]

\[1)\ ABCD - квадрат:\]

\[AC = \sqrt{AD^{2} + CD^{2}} =\]

\[= \sqrt{a^{2} + a^{2}} = a\sqrt{2};\]

\[BO = AO = \frac{1}{2}AC = \frac{a\sqrt{2}}{2};\]

\[AC\bot BD;\ \ \ \]

\[\angle AOB = 90{^\circ}.\]

\[2)\ В\ \mathrm{\Delta}AOB:\]

\[S_{\text{AOB}} = \frac{1}{2}AO \bullet BO = \frac{1}{2} \bullet \frac{a^{2}}{2} = \frac{a^{2}}{4}.\]

\[3)\ Окружность:\]

\[S_{\text{AOB}} = \frac{\pi R^{2}\alpha}{360{^\circ}} = \frac{\pi \bullet \frac{a^{2}}{2} \bullet 90{^\circ}}{360{^\circ}} = \frac{a^{2}\pi}{8};\]

\[S_{\text{AB}} = \frac{a^{2}\pi}{8} - \frac{a^{2}}{4} = \frac{a^{2}(\pi - 2)}{8}.\]

\[Ответ:\ \ \frac{a^{2}(\pi - 2)}{8}.\]