Let ‘O’ be the centre of the circle and AB be the chord of the circle. Here, d = 40 cm
∴ r = 40/2 = 20 cm
Since OA = OB = AB,
∆OAB is an equilateral triangle.
The angle subtended at the centre by the minor
arc AOB is θ = 60° = (60 x π/180)c = (π/3)c
= l(minor arc of chord AB) = rθ = 20 x π/3)
= 20π/3 cm