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Question

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



A
(3083493) cm2
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B
(3203503) cm2
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C
49 cm2
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D
3083 cm2
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Solution

The correct option is A (3083493) cm2
Given that , radius of circle (r ) = 14 cm

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

In Δ AOB
OA=OB=Radius of circle
Δ AOB is isosceles.

OAB=OBA=θ

Now, in Δ OAB
AOB+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. So, ΔAOB is an equilateral triangle.

Also, area of  ΔAOB
=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
=3083 cm2

The area of the minor segment 
= Area of sector OBAO - Area of the equilateral triangle
=(3083493) cm2

Mathematics

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