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Magnitudes:| ec{OA}| = \sqrt{(-1 + \sqrt{7})^2 + (1 + \sqrt{7})^2} = \sqrt{(1 - 2\sqrt{7} + 7) + (1 + 2\sqrt{7} + 7)} = \sqrt{16 + 8} = \sqrt{24} = 2\sqrt{6}| ec{OB}| = \sqrt{(-1 - \sqrt{7})^2 + (1 - \sqrt{7})^2} = \sqrt{(1 + 2\sqrt{7} + 7) + (1 - 2\sqrt{7} + 7)} = \sqrt{16 + 8} = \sqrt{24} = 2\sqrt{6}\cos heta = rac{-12}{(2\sqrt{6})(2\sqrt{6})} = rac{-12}{24} = - rac{1}{2} \Rightarrow heta = rac{2\pi}{3}The area of the circular sector is:A_{ ext{sector}} = rac{1}{2} r^2 heta = rac{1}{2} \cdot 16 \cdot rac{2\pi}{3} = rac{16\pi}{3}Area of triangle \(OAB\):A_{ ext{triangle}} = rac{1}{2} ab \sin heta = rac{1}{2} (2\sqrt{6})(2\sqrt{6}) \sin\left( rac{2\pi}{3}ight) = rac{1}{2} \cdot 24 \cdot rac{\sqrt{3}}{2} = 6\sqrt{3}Area of segment (the smaller region bounded by the chord and arc) is: