Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of `60^(@)` . (use `pi = 3.14`)

Explanation

Given that, radius of a circle (r ) = 12 cm
and central angle of sector `OBCA (theta) = 60^(@)`
`:.` Area of sector `OBCA = (pi r^(2))/(360)xxtheta` [here, OBCA = sector and ABCA = segment]
= `(3.14xx12xx12)/(360^(@))xx60^(@)`
= `3.14xx2xx12`
= `3.14 xx24 = 75.36 cm^(2)`
Since, `DeltaOAB` is an isosceles triangle.
Let `angleOAB = angleOBA = theta_(1)`
and OA = OB 12 cm
`angleAOB = theta= 60^(@)`
`:. angleOAB + angleOBA + angleAOB = 180^(@)` [`because` sum of all interior angles of a triangle is `180^(@)`]
`rArr theta_(1) + theta_(1) + 60^(@) = 180^(@)`
`rArr 2theta_(1) = 120^(@)`
`rArr theta_(1) = 60^(@)`
`:. theta_(1) = theta = 60^(@)`
So, the required `DeltaAOB` is an equilateral triangle.
Now, area of `DeltaAOB=sqrt(3)/(4)("side")^(2)`[`because` area of an equilateral triangle=`sqrt(3)/(4)("side")^(2)`]
= `sqrt(3)/(4)(12)^(2)`
= `sqrt(3)/(4)xx12xx12= 36sqrt(3) cm^(2)`
Now, area of the segment of a circle i.e.,
ABCA = Area of sector OBCA - Area of `DeltaAOB`
`(75.36 - 36sqrt(3)) cm^(2)`
Hence, the required area of segment of a circle is `(75.36 - 36sqrt(3)) cm^(2)` .