In odot(O, 12), minor widehat{ACB} subtends a | vedclass

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(D) Given, the radius of the circle $r = 12 	ext{ cm}$ and the central angle subtended by the minor arc $\widehat{ACB}$ is $	heta = 30^{\circ}$.The

Vedclass · Accepted Answer

$22$

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(D) Given, the radius of the circle $r = 12 	ext{ cm}$ and the central angle subtended by the minor arc $\widehat{ACB}$ is $	heta = 30^{\circ}$.The circumference of the circle is $C = 2\pi r = 2 	imes \pi 	imes 12 = 24\pi 	ext{ cm}$.The length of the minor arc $\widehat{ACB}$ is given by the formula $L_{	ext{minor}} = \frac{	heta}{360^{\circ}} 	imes 2\pi r$.$L_{	ext{minor}} = \frac{30^{\circ}}{360^{\circ}} 	imes 24\pi = \frac{1}{12} 	imes 24\pi = 2\pi 	ext{ cm}$.The length of the major arc $\widehat{ADB}$ is the total circumference minus the length of the minor arc.$L_{	ext{major}} = 24\pi - 2\pi = 22\pi 	ext{ cm}$.

In $\odot(O, 12)$, minor $\widehat{ACB}$ subtends an angle of measure $30^{\circ}$ at the centre. Then, the length of major $\widehat{ADB}$ is $\ldots \ldots \ldots \text{ cm}$. (in $\pi$)

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