(a) Let the angle subtended at the centre by XY = \(\theta\) Length of chord = \(2r \sin \frac{\theta}{2}\) \(\implies 6 cm = 2 \times 5 \times \frac{\theta}{2}\) \(\sin \frac{\theta}{2} = \frac{6}{10} = 0.6\) \(\frac{\theta}{2} = \sin^{-1} (0.6) = 36.87^\circ\) \(\theta = 2 \times 36.87^\circ = 73.74^\circ\) (b) Area of sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) = \(\frac{73.74}{360} \times \frac{22}{7} \times 5^{2}\) = \(16.094 cm^{2}\) Area of triangle = \(\sqrt{s(s - a)(s - b)(s - c)}\) where \(s = \frac{a + b + c}{2}\) \( a = 5 cm ; b = 5 cm ; c = 6 cm \therefore s = \frac{5 + 5 + 6}{2} = \frac{16}{2} = 8\) \(\therefore Area = \sqrt{8(8 - 5)(8 - 5)(8 - 6)}\) = \(\sqrt{8(3)(3)(2)}\) = \(\sqrt{144}\) = \(12 cm^{2}\) Area of shaded portion = \(16.094 cm^{2} - 12 cm^{2} = 4.094 cm^{2}\)