Q) Find the area of the minor and the major sectors of a circle with radius 6 cm, if the angle subtended by the minor arc at the centre is 60 deg (Use pi = 3 * 14 )

Ans: Let’s draw the diagram for better understanding:

Step 1: Here, we are given that θ  = 600, r = 6 cm

∵ ∠ AOB is 600

Therefore, sum of other two angles:

∠ OAB + ∠ OBA = 1800 – 60

= 1200

Now, OA and OB are radius of the circle, hence these are equal.

Therefore, angles opposite to equal sides will also be equal.

∴ ∠ OAB = ∠ OBA

∴ ∠ OAB = ∠ OBA = = 600

Therefore, Δ OAB is an equilateral triangle.

Step 2: Area of minor segment:

From the above diagram, Area of minor segment APB

= Area of sector AOBP – Area of triangle AOB

We know that Area of minor segment APB =

Here, we are given that θ  = 600, r = 6 cm

∴ Area of minor segment APB =

= (3.14) (36) =  18.84 cm2

Next, Area of equilateral Δ AOB =

Here, a = 6 cm, Therefore, area of Δ AOB =

= (36) = 15.57 cm2

Area of minor segment APB = Area of sector AOBP – Area of triangle AOB

= 18.84 – 15.57 = 3.27 cm2

Therefore, the area of minor segment is 3.27 cm2

Step 3: Area of major segment:

From the above diagram, Area of major segment AQB

= Area of the circle  – Area of minor segment APB

= π (6)2 – 3,27 = 3.14 x 36 – 3.27

= 113.04 – 3.27 = 109.77

Therefore, the area of major segment is 109.77 cm2

Please do press “Heart” button if you liked the solution.

Scroll to Top