Question

A chord AB of circle, of radius 14cm makes an angle of 60° at the centre. Find the area of minor segment of circle.

Answer 1

Given radius (r) = 14cm = OA = OB

𝜃 = angle at centre = 60°

In ΔAOB, ∠A = ∠B     [angles opposite to equal sides OA and OB] = x

By angle sum property ∠A + ∠B + ∠O = 180°

x + x + 60° = 180° ⇒ 2x = 120° ⇒ x = 60°

All angles are 60°, OAB is equilateral OA = OB = AB

Area of segment = area of sector – area Δle OAB

=`theta/360^@× pir^2 − sqrt(3)/4× (−AB)^2`

=`60/360×22/7× 14 × 14 − sqrt(3)/4× 14 × 14`

=`308/3− 49sqrt(3)`

`= (308−147sqrt(3))/3cm^2`