**Q) ****An arc of a circle of radius 21 cm subtends an angle of 60° at the centre. Find:****(i) the length of the arc****(ii) the area of the minor segment of the circle made by the corresponding chord radius r of in-circle.**

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

**(i) length of the arc:**

We know that the length of the arc is given by =

Here, we are given that θ = 60^{0}, r = 21 cm

Therefore, length of the arc =

= = 22 cm

**Therefore the length of the arc is 22 cm**

**(ii) 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 = 21 cm

∴ Area of minor segment APB =

= = 11 x 21 = **231 cm ^{2}**

Next Area of Δ AOB:

Here, we can see that Δ AOB is a equilateral triangle.

* (∠ AOB is 60 ^{0 }Therefore, sum of other two angle sis 120^{0 }. Since sides OA and OB are equal, hence angles opposite to equal sides will also be equal. It makes all angles 60^{0})*

Sine area of a equilateral triangle is =

Here, a = 21 cm, Therefore, area of Δ AOB = cm^{2}

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

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