In a circle with centre P,the distance of a c | vedclass

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(A) Let $P$ be the centre of the circle and $AB$ be the chord. Let $M$ be the midpoint of the chord $AB$ such that $PM \perp AB$. Given that the dista

Vedclass · Accepted Answer

$52$

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(A) Let $P$ be the centre of the circle and $AB$ be the chord. Let $M$ be the midpoint of the chord $AB$ such that $PM \perp AB$. Given that the distance of the chord from the centre is $PM = 20 \, cm$. The length of the chord $AB = 96 \, cm$. Since the perpendicular from the centre to a chord bisects the chord,we have $AM = MB = \frac{AB}{2} = \frac{96}{2} = 48 \, cm$. In the right-angled triangle $	riangle PMA$,by the Pythagorean theorem,$PA^2 = PM^2 + AM^2$. Here,$PA$ is the radius $(r)$ of the circle. $r^2 = 20^2 + 48^2$. $r^2 = 400 + 2304 = 2704$. $r = \sqrt{2704} = 52 \, cm$. Thus,the radius of the circle is $52 \, cm$.

In a circle with centre $P$,the distance of a chord $AB$ from the centre $P$ is $20 \, cm$. If $AB = 96 \, cm$,then the radius of the circle is .......... $cm$.

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