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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.

Answers (1)


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 = ?)

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   \DeltaOAB is equilateral triangle
So, each angle equilateral is 60\degree
\therefore \theta = 60\degree  = \frac{\pi}{3}radian
Now, we have \theta and r 
           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

Posted by

Safeer PP

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