Question

in a circle of radius 21 cm, an are subtends an angle of 60 at the centre. Find:
(1) the length of the arc (2) area of the sector formed by the are
(3) area of the segment formed by the corresponding chord​

Answer 1

Answer:

hope this was helpful...

Answer 2

Step-by-step explanation:

Given:-

In a circle of radius 21 cm, an are subtends an angle of 60 at the centre.

To find:-

Find:

(1) the length of the arc

(2) area of the sector formed by the arc

(3) area of the segment formed by the

corresponding chord?

Solution:-

Radius of the circle (r)=21 cm

Angle subtended by an arc at the centre (X°) = 60°

Length of the arc (l) = (X°/360°)×2πr units

=>l = (60°/360°)×2×(22/7)×21 cm

=>l = (1/6)×2×22×3 cm

=>l=2×22×3/6

=>l = 6×22/6

=>l = 22

=>l = 22 cm

We know that

Area of a sector = (A)=lr/2 sq.units

=>A = 22×21/2 sq.cm

=>A = 11 ×21 sq.cm

A = 231 sq.cm

Area of a segment formed by the corresponding arc = Area of a sector - Area of a triangle

=>231- (1/2)r^2 Sin X°

=>231 - (1/2)×(21)^2× Sin 60°

=>231 - (21×21×√3)/2×2

=>231 - 441√3/4

=>[4(231)-441√3]/4

=>(924-441√3)/4

We know that √3 = 1.732 then

=>[924-441(1.732)]/4

=>(924-763.81)/4

=>160.19/4

=>40.05 sq.cm

Answer:-

1)the length of the arc = 22 cm

2)area of the sector formed by the arc

= 231 sq.cm

3)area of the segment formed by the corresponding chord = 40.05 sq.cm

Used formulae:-

• Length of the arc (l) = (X°/360°)×2πr units
• Area of a sector = (A)=lr/2 sq.units
• Area of a sector = (X°/360°)×πr^2 sq.units
• Area of a segment formed by the corresponding arc = Area of a sector - Area of a triangle
• Area of a triangle = (1/2)r^2 sin X°