**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**