in a circle of radius 21cm,an arc subtends an angle of 60° at the centre. Find area of the segment formed by the corresponding chord.
Answers
Answer:
Area of segment APB = Area of sector OAPB - Area of AOAB
(231-441/4√3)cm^3
Step-by-step explanation:
Given:-
In a circle of radius 21cm,an arc subtends an angle of 60° at the centre.
To find:-
Find area of the segment formed by the corresponding chord?
Solution:-
Radius of a circle (r)=21 cm
Angle subtends at the centre of the circle by a chord (X°)=60°
Area of a sector = (X°/360°)×πr^2 sq.units
=> (60°/360°)×(22/7)×21^2 sq.cm
=> (1/6)×(22/7)×21×21
=> (1×22×21×21)/(6×7)
=> 9702/42
=> 231 sq.cm
Area of the sector = 231 sq.cm -------(1)
Are of a triangle of the angle 60° and the radius r 21 cm
=> (1/2)r^2 Sin X°
=> (1/2)×21^2×Sin 60°
=> (1/2)×21×21×(√3/2)
=> (1×21×21×√3)/(2×2)
=> 441√3/4 sq.cm
We know that √3 = 1.732 then
=> 441×1.732/5
=> 763.812/4
=> 190.953 sq.cm
=> 190.95 sq.cm
(correct it upto two decimals)
Area of the triangle formed by the chord in tje sector = 190.95 sq.cm ---------(2)
We know that
Area of a segment formed by the chord
= Area of a sector - Area of a triangle
From (1) and (2)
=231 sq.cm - 190.95 sq.cm
= 40.05 sq.cm
Area of the segment = 40.05 sq.cm
Answer:-
Area of the segment formed by the chord in the given circle is 40.05 sq.cm
Used formulae:-
- Area of a sector = (X°/360°)×πr^2 sq.units
- Are of a triangle=(1/2)r^2 Sin X°
- Area of a segment formed by the chord = Area of a sector - Area of a triangle
- X° = angle subtended by the chord
- r=radius of the circle
- π=22/7
- √3 = 1.732