**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 = 60^{0}, r = 6 cm

∵ ∠ AOB is 60^{0 }

Therefore, sum of other two angles:

∠ OAB + ∠ OBA = 180^{0} – 60^{0 }

= 120^{0 }

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 = = 60^{0}

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 = 60^{0}, r = 6 cm

∴ Area of minor segment APB =

= (3.14) (36) = ** 18.84 cm ^{2}**

Next, Area of equilateral Δ AOB =

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

= (36) = **15.57 cm ^{2}**

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

= 18.84 – 15.57 = 3.27 cm^{2}

**Therefore, the area of minor segment is 3.27 cm ^{2}**

**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 cm ^{2}**

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