P is the centre of the circle of radius 20, c | vedclass

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Question

(C) From $P$,draw $PM \perp AB$. Join $PA$.The perpendicular from the centre of a circle to a chord bisects the chord.$	herefore AM = MB = \frac{1}{2

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) From $P$,draw $PM \perp AB$. Join $PA$.The perpendicular from the centre of a circle to a chord bisects the chord.$	herefore AM = MB = \frac{1}{2} AB = \frac{1}{2} 	imes 32 = 16\, cm$.$PA$ is the radius of the circle,so $PA = 20\, cm$.In $\Delta PMA$,$\angle M = 90^{\circ}$.By Pythagoras theorem,$PA^2 = PM^2 + AM^2$.$	herefore PM^2 = PA^2 - AM^2 = (20)^2 - (16)^2 = 400 - 256 = 144$.$	herefore PM = \sqrt{144} = 12\, cm$.Thus,the distance of the chord $AB$ from the centre $P$ is $12\, cm$.

$P$ is the centre of the circle of radius $20\, cm$. $AB$ is a chord of the circle. If $AB = 32\, cm$,then find the distance of the chord $AB$ from the centre $P$.

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