(a) Area of the shaded segment = area of sector XOY - area of triangle OXY. \(\implies 504 = \frac{\theta}{360^\circ} \times \pi r^{2} - \frac{1}{2} r^{2} \sin \theta\) \(\implies 504 = \frac{90}{360} \times \frac{22}{7} \times r^{2} - \frac{1}{2} \times r^{2} \sin 90\) \(504 = r^{2}(\frac{22}{28} - \frac{1}{2})\) \(504 = \frac{2r^{2}}{7} \implies r^{2} = \frac{504 \times 7}{2} = 1764\) \(r = \sqrt{1764} = 42 cm\). (b) In the diagram, \(a_{1} = a_{2}\) (base angles of an isosceles triangle) \(\therefore a_{1} + a_{2} = 2a_{1}\) \(\therefore 2a_{1} + 109^\circ = 180^\circ \implies 2a_{1} = 180^\circ - 109^\circ = 71^\circ\) \(a_{1} = a_{2} = \frac{71^\circ}{2} = 35.5^\circ\) Hence, \(< RQS = 66^\circ + 35.5^\circ\) = \(101.5^\circ\)