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Question

Question 10
Find the area of the minor segment of a circle of radius 14 cm , when the angle of the corresponding sector is
60

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Solution

Given that , radius of circle (r ) = 14 cm

Angle of the corresponding sector. i.e central angle (θ)=60

Since, in ΔAOB, OA=OB=Radius of circle. i.e., ΔAOB is isosceles.

OAB=OBA=θ

Now, in ΔOABAOB+OAB+OBA=180

[ Since , sum of interior angles of any triangle is 180 ]

60+θ+θ=180 [given,AOB=60]

2θ=120

θ=60

i.e AOB=OBA=60=AOB

Since, all angles of ΔAOB are equal to 60 i.e ΔAOB is an equilateral triangle.

Also, OA=34(side)2

=34×(14)2 [Area of an equilateral triangle=34(side)2]

34×196=493 cm2

Area of sector OBAO =πr2360×θ

=227×14×14360×60

=22×2×146=22×143=3083cm2

The area of the minor segment
= Area of sector OBAO - Area of the equilateral triangle

(3083493)cm2






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