Solution

Diameter of the circle $$= 40$$ cm
$$\therefore$$ Radius (r) of the circle $$\displaystyle=\frac{40}{2}=20$$ cm
Let AB be a chord (length = 20 cm) of the circle.
In $$\triangle{OAB}$$, OA = OB = Radius of circle = 20 cm
Also, AB = 20 cm
Thus, $$\triangle{OAB}$$ is an equilateral triangle.

$$\displaystyle\therefore\theta=60^\circ=\frac{\pi}{3}$$ radian
We know that in a circle of radius r unit, if an arc of length I unit subtends an angle $$\theta$$ radian at the centre, then $$\displaystyle\theta=\frac{l}{r}$$
$$\therefore \displaystyle\frac{\pi}{3}=\frac{\displaystyle {arc\ AB}}{20}\implies {arc\ AB}=\frac{20}{3}\pi$$ cm