AREAS RELATED
TO CIRCLE
Area of a sector and segment
60
360
×
2
×
22
7
×
21
Q. In a circle of radius 21 cm, an arc subtends
an angle of 60° at the centre. Find :
(i) The length of the arc
l (arc AXB) =
=
= 22 cm
(ii) Area of the sector formed by the arc
A
B
X
21 cm
O
corresponding chord.
∴
60°
Sol.
6
3
2
θ
360
×
2πr
60
360
×
22
7
×
21
×
21
A (O – AXB) =
=
= 231 cm2
∴
6
3
11
7
θ
360
×
πr2
21 cm
What is formula to find length of arc?
What is formula to find area of sector?
Length of arc is 22 cm.
∴
Area of sector (O – AXB) is 231 cm2.
∴
∴
In ΔOAB,
OA
∠A
∠A
180º
x
180
∴
2x =
180
∴
x =
60
∴
Sol.
[radius of same circle]
OB
[Angles opposite to equal sides are equal]
x
=
∠B
=
60
=
=
∠O
+
∠B
+
x
120
2
x =
=
∴
+
+
ar (ΔOAB) =
=
=
ΔOAB is an equilateral triangle
∴
4
×
(Side)2
4
×
21
×
21
4
cm2
ar (ΔOAB)
Q. In a circle of radius 21 cm, an arc subtends
an angle of 60° at the centre. Find :
(ii) Area of the sector formed by the arc
corresponding chord.
(i) The length of the arc
Area of minor segment =
A
B
X
21 cm
O
21 cm
ar(O – AXB) –
✔
?
x
x
We know, sum of measures of angles of triangle is 180o
In ΔOAB, each angle is 60o
∴ ΔOAB is equilateral triangle
What is formula to find area of equilateral triangle ?
4
×
(Side)2
60°
60
2x =
120
60
60°
60°
ar(ΔOAB)
✔
–
ar(O – AXB)
=
231 cm2
Area of minor segment =
Area of Minor Segment =
∴
ar(O – AXB) – ar(ΔOAB)
231
–
4
cm2
441
Sol.
Q. In a circle of radius 21 cm, an arc subtends
an angle of 60° at the centre. Find :
(ii) Area of the sector formed by the arc
(i) The length of the arc
corresponding chord.
A
B
X
21 cm
O
21 cm
60°
60°
60°
Area of minor segment =
ar(O – AXB) –
ar(ΔOAB)
Area of Minor Segment is
∴
231
–
4
cm2
441
ar(O – AXB)
=
231 cm2
ar(ΔOAB)
=
4
cm2