Solution:
From the given information, Radius of the circle = r = \[12\] cm
∴ OA = OB = \[12\] cm
\[\angle AOB={{60}^{\circ }}\] (given)
As triangle OAB is an isosceles triangle, ∴ \[\angle OAB=\angle OBA=\theta \] (say)
Also, we know that Sum of interior angles of a triangle is 180°,
∴ \[\theta +\theta +{{60}^{\circ }}={{180}^{\circ }}\]
⇒ \[2\theta ={{120}^{\circ }}\]
⇒ \[\theta ={{60}^{\circ }}\]
Therefore, triangle AOB is an equilateral triangle.
∴ AB = OA = OB = \[12\] cm
Area of the triangle AOB of side a= \[(\sqrt{3}/4)\times {{a}^{2}}\]
= \[(\sqrt{3}/4)\times {{(12)}^{2}}\]
= \[(\sqrt{3}/4)\times 144\]
= \[36\sqrt{3}\] \[c{{m}^{2}}\]
= \[62.354\]\[c{{m}^{2}}\]
Now, for Central angle of the sector AOBCA = \[\phi ={{60}^{\circ }}=(60\pi /180)=(\pi /3)\]radians
Thus, we know that area of the sector AOBCA = \[\frac{1}{2}{{r}^{2}}\phi \]
= \[\frac{1}{2}\times {{12}^{2}}\times \pi /3\]
= \[{{12}^{2}}\times (22/(7\times 6))\]
= \[75.36\]\[c{{m}^{2}}\]
Now, for Area of the segment ABCA = Area of the sector AOBCA – Area of the triangle AOB
= \[(75.36-62.354)\]\[c{{m}^{2}}\] = \[13.006\] \[c{{m}^{2}}\]