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

1 Answer

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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