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.
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.