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(A) Diameter of the circle $= 40 \, cm$.Radius $(r)$ of the circle $= \frac{40}{2} \, cm = 20 \, cm$.Let $AB$ be a chord of length $20 \, cm$ in the c

Vedclass · Accepted Answer

$\frac{20 \pi}{3} \, cm$

Vedclass · Answer

(A) Diameter of the circle $= 40 \, cm$.Radius $(r)$ of the circle $= \frac{40}{2} \, cm = 20 \, cm$.Let $AB$ be a chord of length $20 \, cm$ in the circle with center $O$.In $\Delta OAB$,$OA = OB = r = 20 \, cm$ and $AB = 20 \, cm$.Since all sides are equal,$\Delta OAB$ is an equilateral triangle.Therefore,the central angle $	heta = 60^{\circ} = \frac{\pi}{3}$ radians.The length of the arc $(l)$ is given by the formula $l = r 	heta$.$l = 20 	imes \frac{\pi}{3} = \frac{20 \pi}{3} \, cm$.Thus,the length of the minor arc is $\frac{20 \pi}{3} \, cm$.

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

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