(b) \(\bigg[a^2+4\bigg[rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg]\bigg]\)As shown in the given figures, if ‘a’ is each side of the square, then ∠DOC = 120º ⇒ ∠ODC = ∠OCD = 30ºNow in fig. (iii), \(rac{PC}{OC}\) = sin 60º ⇒ \(rac{rac{a}{2}}{OC}\) = \(rac{\sqrt3}{2}\)⇒ OC = \(rac{a}{\sqrt3}\), i.e., radius of arc CDAlso, OP = \(rac{a}{2}\) cot 60º = \(rac{a}{2\sqrt3}\)Now are of ΔODC = \(rac{1}{2}\)x CD x OP = \(rac{1}{2}\) x a x \(rac{a}{2\sqrt3}\) = \(rac{a^2}{4\sqrt3}\)Area of sector DOC = \(\pi imes\bigg(rac{a}{\sqrt3}\bigg)^2 imesrac{120}{360}=rac{\pi{a}^2}{9}\)[Using Area of secotr = \(\pi{R}^2\big(rac{ heta}{360°}\big)\)]∴ Area of segment DLC = Area of sector DOC – Area of ΔDOC= \(rac{\pi{a}^2}{9}\) - \(rac{a^2}{4\sqrt3}\)∴ Total area of the figure = Area of square + Total area of 4 segments = \(a^2+4\bigg(rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg).\)