Table of contents  
Important Formulas related to Circle  
Minor Arc and Major Arc  
Sector of a Circle and its Area  
Segment of a Circle and its Area 
An arc length is called a major arc if the arc length enclosed by the two radii is greater than a semicircle.
If the arc subtends angle ‘θ’ at the centre, then the
Length of minor arc =
Length of major arc =
A region of a circle is enclosed by any two radii and the arc intercepted between two radii is called the sector of a circle.
A sector is called a minor sector if the minor arc of the circle is part of its boundary.
OAPB is minor sector.
Area of minor sector =
Perimeter of minor sector =
A sector is called a major sector if the major arc of the circle is part of its boundary.
OAQB is major sector
Area of major sector =
Perimeter of major sector =
The region enclosed by an arc and a chord is called a segment of the circle. The region enclosed by the chord AB & minor arc ACB is called the minor segment.
Area of Minor segment = Area of the corresponding sector – Area of the corresponding triangle
The region enclosed by the chord AB & major arc ADB is called the major segment. Area of major segment = Area of a circle – Area of the minor segment
Area of major sector + Area of triangle
Example: A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73).
Sol:
We use the concept of areas of sectors of circles to solve the question.
In a circle with radius r and the angle at the centre with degree measure θ:
(i) Area of the sector = θ/360 πr2
(ii) Area of the segment = Area of the sector  Area of the corresponding triangle
Let's draw a figure to visualize the given question.
Here, radius, r = 12 cm, ∠AOB = θ = 120°
Visually it’s clear from the figure that AB is the chord that subtends 120° angle at the centre.
To find the area of the segment AYB, we have to find the area of the sector OAYB and the area of the ΔAOB
(i) Area of sector OAYB = θ/360° πr^{2}
(ii) Area of ΔAOB = 1/2 × base × height
For finding the area of ΔAOB, draw OM ⊥ AB then find base AB and height OM using the figure as shown above.
Area of sector OAYB = 120°/360° × πr^{2}
= 1/3 × 3.14 × (12 cm)^{2}
= 150.72 cm^{2}
Draw a perpendicular OM from O to chord AB
In ΔAOM and ΔBOM
AO = BO = r (radii of circle)
OM = OM (common side)
∠OMA = ∠OMB = 90° (perpendicular OM drawn)
∴ ΔAOM ≅ ΔBOM (By RHS Congruency)
⇒ ∠AOM = ∠BOM (By CPCT)
Therefore, ∠AOM = ∠BOM = 1/2 ∠AOB = 60°
In ΔAOM,
AM/OA = sin 60° and OM/OA = cos 60°
AM/12 cm = √3/2 and OM/12 cm = 1/2
AM = √3/2 × 12 cm and OM = 1/2 × 12 cm
AM = 6√3 cm and OM = 6 cm
⇒ AB = 2 AM
= 2 × 6√3 cm
= 12√3 cm
Area of ΔAOB = 1/2 × AB × OM
= 1/2 × 12√3 cm × 6 cm
= 36 × 1.73 cm^{2}
= 62.28 cm^{2}
Area of segment AYB = Area of sector OAYB  Area of ΔAOB
= 150.72 cm^{2}  62.28 cm^{2}
= 88.44 cm^{2}
^{}
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