A chord of a circle of radius 21 cm subtends an angle of 60° at the center. Find the area of the corresponding minor segment of the circle. [Use π = 22/7 and √3 = 1.73].

Radius of the circle = r = 21 cm

∴ OA = OB = 21 cm

∠ AOB = 60° (given)

Since triangle OAB is an isosceles triangle, ∴ ∠OAB = ∠OBA = θ (say)

Also, Sum of interior angles of a triangle is 180°,

∴ θ + θ + 60° = 180°

⇒2θ = 120° ⇒ θ = 60°

Thus the triangle AOB is an equilateral triangle.

∴ AB = OA = OB = 21 cm

Area of the triangle AOB × a², where a is the side of the triangle.

Now, Central angle of the sector AOBA = θ = 60° = 60π/180 = (π/3) radians

Thus, area of the sector AOBA = (1/2)(r²θ)

= 231cm²

Now, Area of segment ABA = Area of sector AOBA – Area of the triangle AOB

= (231 – 190.7325) cm² = 40.2675 cm²