(b) 7√3 cm2.
AB = 6 cm, ∠C = 60º (∴ ∠A = ∠B = 60º)
∴ ΔABC is an equilateral triangle
Area of ΔABC = \(\frac{\sqrt3}{4}\) × (6)2 = 9√3
Area of (ΔADE + ΔBFG) = 2 x \(\bigg(\frac{\sqrt3}{4}\times(2)^2\bigg)\) = 2√3
∴ Area of pentagon = 9√3 - 2√3 = 7√3 cm2.