(a) Area of the shaded segment = Area of sector - Area of triangle = \(\frac{\theta}{360^\circ} \times \pi r^{2} - \frac{1}{2} ab \sin \theta\) Area of sector = \(\frac{63}{360} \times \frac{22}{7} \times 10^{2}\) = \(55 cm^{2}\) Area of triangle = \(\frac{1}{2} \times 10 \times 10 \times \sin 63^\circ\) = \(50 \times 0.891 = 44.55 cm^{2}\) \(\therefore \text{Area of the shaded segment} = 55 - 44.55 = 10.45 cm^{2}\) (b) (i) Total surface area = \(2\pi r (r + h)\) = \(2 \times \frac{22}{7} \times 3 (3 + 6)\) = \(\frac{132}{7} \times 9 = \frac{1188}{7}\) = \(169.71 cm^{2}\) (ii) Volume of tin = \(\pi r^{2} h\) = \(\frac{22}{7} \times 3 \times 3 \times 6 = \frac{1188}{7}\) \(\approxeq 169.71 cm^{3}\)