What is the area of minor segment of a circle of radius 14 cm when its Centre angle is 60 degrees?

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10 Questions 10 Marks 6 Mins

Given:

Radius of circle = 14 cm

Central angle = 60°

Formula used:

Area of minor segment = \(πr^2 × \frac{θ}{360°} - \frac{1}{2}r^2sinθ\)

where r is the radius

and θ is the angle of the segment

Sin 60° = \(\frac{\sqrt3}{2}\)

Calculation:

According to the question,

Area of segment = \(\frac{22}{7} × 14^2 × \frac{60°}{360°} - \frac{1}{2} × 14^2sin60°\)

⇒ \(22× 28× \frac{1}{6} - 98× \frac{\sqrt3}{2}\)

⇒ 102.67 - 49 × 1.73

⇒ 102.67 - 84.87 = 17.79 ≈ 17.8 cm2

∴ The area of minor segment is 17.8 cm2

Alternate Method

Given:

Radius of circle = 14 cm

Central angle = 60°

Formula used:

Area of the circle = πr2

Area of sector = \(πr^2 × \frac{θ}{360°}\) 

Calculation:

According to the diagram,

Area of the circle = πr2

⇒ 22/7 × (14)2 = 616 cm2

Area of the sector AOB = \(πr^2 × \frac{θ}{360°}\)

⇒ \(\frac{22}{7} × 14^2 × \frac{60°}{360°}\)

⇒ 616 × 1/6 = 102.67 cm2

Area of Δ AOB = \(\frac{\sqrt3}{4} \times (AO)^2\)

⇒ Area of the minor segment = Area of the sector AOB - Area of Δ AOB

⇒ 102.67 - 84.87 = 17.79 ≈ 17.8 cm2

∴ The area of minor segment is 17.8 cm2

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