# Q (5) In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

Solution

Given :- radius (r)of circle = $\frac{Diameter}{2} = \frac{40cm}{2} = 20 cm$

length of chord = 20 cm

We know that
$\theta = \frac{l}{r}$                                      (r = 20cm , l = ? , $\theta$ = ?)

Now,

AB is the chord of length 20cm and OA and OB are radii of circle 20 cm each
The angle subtended by OA and OB at centre = $\theta$
$\because$  OA = OB = AB
$\therefore$   $\Delta$OAB is equilateral triangle
So, each angle equilateral is $60\degree$
$\therefore$ $\theta = 60\degree$  $= \frac{\pi}{3}radian$
Now, we have $\theta$ and r
So,
$l = r\theta = 20\times\frac{\pi}{3}=\frac{20\pi}{3}$
$\therefore$ the length of the minor arc of the chord (l) = $\frac{20\pi}{3}$  cm

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