Q) A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle. Also find the area of the major segment of the circle.

Ans: 

Let the chord AB cut the circle in 2 parts. Sector APB is minor segment and AQB is major one.

Now in △AOB, radius OA=OB=14 cm

Therefore, ∠OAB  = ∠OBA  [identity: angles opposite to equal side]

Given that ∠AOB=60°

Therefore, ∠AOB + ∠OAB + ∠OBA = 180°

or  60° + 2 ∠OAB = 180°

or 2 ∠OAB = 120°

∠OAB = 60° and ∠OBA = 60°

Since All angles are 60°, therefore △OAB is an equilateral triangle

Now, area of minor segment APB = Area of sector OAPB – Area of △OAB

= πr²   − (OA)²                     

[To understand Area of sector of a circle, click here: “Area of sector of circle”]

[To understand Area of an equilateral triangle, click here: “Area of equilateral triangle”]

=  x 14 x 14 x   − x 14 x 14                       

= – 49√3

= 102.667 – 84.868 = 17.799 cm²

Now, Area of major segment AQB = Area of circle – Area of minor segment APB

= π r² − Area of segment APB                       

= x 14 x 14 – 17.799

= 22 x 2 x 14 – 17.799

= 616 – 17.799

= 598.201 cm²

Hence, the area of minor segment is 17.799 cm² and area of the major segment of the circle is 598.201 cm².