Find the area of the segment of a circle of radius \[12\] m whose corresponding sector has a central angle of \[{{60}^{\circ }}\] (Use \[\pi =3.14\]).

Home » Find the area of the segment of a circle of radius m whose corresponding sector has a central angle of (Use ).

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

$\pi =3.14$

).

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

=

$(75.36-62.354)$

$c{{m}^{2}}$

=

$13.006$

$c{{m}^{2}}$

i

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