AD is a diameter of a circle and AB is a chor | vedclass

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(C) Let $O$ be the centre of the circle. Draw $OP \perp AB$.Since the perpendicular from the centre to a chord bisects the chord,we have:$AP = \frac{1

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) Let $O$ be the centre of the circle. Draw $OP \perp AB$.Since the perpendicular from the centre to a chord bisects the chord,we have:$AP = \frac{1}{2} 	imes AB = \frac{1}{2} 	imes 30 = 15 \, cm$.The radius of the circle is $OA = \frac{AD}{2} = \frac{34}{2} = 17 \, cm$.In the right-angled triangle $\Delta OPA$,by the Pythagorean theorem:$OP^2 + AP^2 = OA^2$$OP^2 + 15^2 = 17^2$$OP^2 + 225 = 289$$OP^2 = 289 - 225 = 64$$OP = \sqrt{64} = 8 \, cm$.Thus,the distance of the chord $AB$ from the centre is $8 \, cm$.

$AD$ is a diameter of a circle and $AB$ is a chord. If $AD = 34 \, cm$ and $AB = 30 \, cm$,the distance of $AB$ from the centre of the circle is (in $cm$):

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