Find the area of the minor segment of a circle of radius 14cm, when the angle of the corresponding sector is `60^(@)`

Explanation

Given that, radius of circle ( r) = 14 cm
and angle of the corresponding sector i.e., central angle `(theta) = 60^(@)`
Since, in `DeltaAOB` OA = OB= Radius of circle i.e., `DeltaAOB` is isosceles.
`rArr angleOAB = angleOBA = theta`
Now, in `DeltaOAB angleAOB + angleOBA + angleOBA = 180^(@)`
[since, sum of interior angles of any triangle is `180^(@)` ]
`rArr 60^(@) + theta + theta = 180^(@)` [given, `angleAOB = 60^(@)`]
`rArr 2theta = 120^(@)`
`rArr theta = 60^(@)`
i.e., `angleOAB = angleOBA = 60^(@)= angleAOB`
Since, all angles of `DeltaAOB` are equal to `60^(@)` i.e., `DeltaAOB` is an equilateral triangle.
Also, OA = OB = AB = 14 cm
So, Area of `DeltaOAB = sqrt(3)/(4)(side)^(2)`
= `sqrt(3)/(4)xx(14)^(2)` [`:.`area of an equilateral triangle= `sqrt(3)/(4)("side")^(2)`]
`sqrt(3)/(4)xx196= 49sqrt(3)cm^(2)`
and area of sector OBAO= `(pir^(2))/(360^(2))xxtheta`
= `(22)/(7)xx(14xx14)/(360)xx60^(@)`
= `(22xx2xx14)/(6)= (22xx14)/(3)= (308)/(3)cm^(2)`
`:.` Area of minor segment = Area of sector `OBAO - "Area of" DeltaOAB`
= `(308/(3)-49sqrt(3))cm^(2)`
Hence, the required area of the minor segment is `((308)/(3)-49sqrt(3))cm^(2)`