An arc of a circle of radius 21 cm subtends an angle of 60

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 = $\frac{\theta}{360} (2\pi r)$

Here, we are given that θ = 60⁰, r = 21 cm

Therefore, length of the arc = $\frac{60}{360} [2 (\frac{22}{7}) (21)]$

= $\frac{1}{6} (2)(22)(3)$ = 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 = $\frac{\theta}{360} (\pi r^2)$

Here, we are given that θ = 60⁰, r = 21 cm

∴ Area of minor segment APB = $\frac{60}{360} [\frac{22}{7} (21)^2]$

= $\frac{1}{6} (22)(3) (21)$ = 11 x 21 = 231 cm²

Next Area of Δ AOB:

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

(∠ AOB is 60⁰Therefore, sum of other two angle sis 120⁰. Since sides OA and OB are equal, hence angles opposite to equal sides will also be equal. It makes all angles 60⁰)

Sine area of a equilateral triangle is = $\frac{\sqrt 3}{4} (a)^2$