**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

= πr^{2} − (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 cm^{2}

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

= π r^{2} − Area of segment APB ^{ }

= x 14 x 14 – 17.799

= 22 x 2 x 14 – 17.799

= 616 – 17.799

= 598.201 cm^{2}

**Hence, the area of minor segment is 17.799 cm ^{2 }and area of the major segment of the circle is 598.201 cm^{2}**

**.**