Question

Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60° (Use π = 3.14).

Answer 1

According to the question,

Radius of the circle = r = 12 cm

∴ OA = OB = 12 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 = 12 cm

Area of the triangle AOB = (√3 /4)× a²,where a is the side of the triangle.

= (√3 /4) × (12)²

= (√3 /4) × 144

= 36√3 cm²

= 62.354 cm²

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

Thus, area of the sector AOBCA = ½ r² ∅

= ½ × 12² × π/3

= 12² × (22 / (7×6))

= 75.36 cm²

Now, Area of the segment ABCA = Area of the sector AOBCA – Area of the triangle AOB

= (75.36 – 62.354) cm² = 13.006 cm²