Question

A chord of a circle, of radius 14 cm, subtends an angle of 60° at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

Answer 1

Radius of circle = 14 cm

∠AOB = 60°

Area of the smaller sector = `πr^2 xx θ/(360^circ)`

= `22/7 xx 14 xx 14 xx (60^circ)/(360^circ)`

= `(22 xx 2 xx 14)/6`

= `(22 xx 14)/3`

= `308/3`

⇒ Area of the smaller sector = 102.67 cm²

Now, perimeter of the smaller segment = Length of arc AB + Length of chord AB

= `2πr xx θ/(360^circ) + 14`

= `2 xx 22/7 xx 14 xx (60^circ)/(360^circ) + 14`

= `(2 xx 22 xx 2)/6 + 14`

= `(2 xx 22)/3 + 14`

= `44/3 + 14`

= `(44 + 42)/3`

⇒ Perimeter of the smaller segment = `86/3` = 28.67 cm.