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