a. 2m = 2 × n (the angle at the centre is twice the angle at the circumference) ∴ m = n ∠SOR = ∠QRO = 54° (alternate angles are equal) From ∆ROQ Since |OQ| = |OR| (radii), therefore, ∆ROQ is an isosceles triangle 2m = 180° - 2(54°) = 180° - 108° (sum of angles in a ∆ is 180°) 2m = 72° m = \(\frac{72}{2}\) = 36° Recall: m = n ∴ m = n = 36° b. Let the length of the rectangle be L and the width be W L = W + 4 (given) Perimeter = 2(L + W) 40 = 2(L + W) 40 = 2((W + 4) + W) (Recall: L = W + 4) 40 = 2(2W + 4) 40 = 4W + 8 40 - 8 = 4W 32 = 4W W = \(\frac{32}{4}\) W = 8cm L = W + 4 = 8 + 4 L = 12cm Area of a rectangle = L x W = 12 × 8 = \(96 cm^2\)