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

$\begin{array}{1 1}(A)\;\large\frac{10\pi}{3}\normalsize cm&(B)\;\large\frac{20}{33}\normalsize cm\\(C)\;\large\frac{20\pi}{3}\normalsize cm&(D)\;\large\frac{12\pi}{5}\normalsize cm\end{array} $

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  • $\theta=\large\frac{l}{r}$
  • $\theta$=angle subtended by arc
  • $l$=length of the arc
  • $r$=radius of the circle
  • $1^{\large\circ}=(\large\frac{\pi}{180})$ radian=0.01746(approx)
  • Diameter=$2\times$ radius
Given :
Length of the chord=20cm
Since radius=length of chord=20cm
Hence the formed $\Delta$le in the circle is equilateral $\Delta$le
$\Rightarrow \theta=60^{\large\circ}$
We know that $l=r\theta$
$l=20\times 60^{\large\circ}\times \large\frac{\pi}{180^{\large\circ}}$
Thus length of the minor arc of the chord is $\large\frac{20\pi}{3}$
Hence (C) is the correct answer.
answered Apr 16, 2014 by sreemathi.v

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