In a circle of diameter $40 \,cm ,$ the length of a chord is $20 \,cm .$ Find the length of minor arc of the chord.

Diameter of the circle $=40 \,cm$

Radius $(r)$ of the circle $=\frac{40}{2} \,cm =20\, cm$

Let $AB$ be a chord (length $= 20$ $cm$ ) of the circle.

In $\Delta O A B, O A=O B=$ Radius of circle $=20 \,cm$

Also, $A B=20\, cm$

Thus, $\Delta O A B$ is an equilateral triangle.

$\therefore \theta=60^{\circ}=\frac{\pi}{3}$ radian

We know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends an angle $\theta$ radian at the centre then

$\theta=\frac{l}{r}$

${\frac{\pi }{3} = \frac{{\widehat {AB}}}{{20}} \Rightarrow \widehat {AB} = \frac{{20\pi }}{3}\,cm}$

Thus, the length of the minor arc of the chord is $\frac{20 \pi}{3} \,cm $