In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord

Asked by Abhisek | 1 year ago |  122

##### Solution :-

Given,

Radius = 21 cm

θ = 60°

(i) Length of an arc = $$\dfrac{ θ}{360°}×Circumference(2πr)$$

Length of an arc AB = $$\dfrac{60°}{360°}×2×\dfrac{22}{7}×21$$

$$\dfrac{1}{6}×2× \dfrac{22}{7}×21$$

Or Arc AB Length = 22cm

(ii) It is given that the angle subtend by the arc = 60°

So, area of the sector making an angle of 60°

$$( \dfrac{60°}{360°})$$×π rcm2

$$\dfrac{441}{6}\times\dfrac{22}{7}$$ cm2

Or, the area of the sector formed by the arc APB is 231 cm2

(iii) Area of segment APB = Area of sector OAPB – Area of ΔOAB

Since the two arms of the triangle are the radii of the circle and thus are equal,

and one angle is 60°, ΔOAB is an equilateral triangle.

So, its area will be $$\dfrac{\sqrt{3}}{4}$$×a2 sq. Units.

Area of segment APB = $$231-(\dfrac{\sqrt{3}}{4})×(OA)^2$$

$$231-(\dfrac{\sqrt{3}}{4})×212$$

Or, Area of segment APB

$$[231-\dfrac{441\sqrt{3}}{4}]$$ cm2

Answered by Pragya Singh | 1 year ago

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