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Find the area of the segment of a circle of radius \[12\] m whose corresponding sector has a central angle of \[{{60}^{\circ }}\] (Use \[\pi =3.14\]).
Find the area of the segment of a circle of radius \[12\] m whose corresponding sector has a central angle of \[{{60}^{\circ }}\] (Use \[\pi =3.14\]).
Areas Related to Circles | Exercise 11.4 | Maths | NCERT Exemplar
Find the area of the segment of a circle of radius \[12\] m whose corresponding sector has a central angle of \[{{60}^{\circ }}\] (Use \[\pi =3.14\]).

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

= \[(75.36-62.354)\]\[c{{m}^{2}}\] = \[13.006\] \[c{{m}^{2}}\]

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