The radius of a circle with centre P is 50 , | vedclass

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(B) Let the radius $r = 50 \, cm$. Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively.Since $AB = 80 \, cm$,$AM = MB = 40 \, cm$.In

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$60$

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(B) Let the radius $r = 50 \, cm$. Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively.Since $AB = 80 \, cm$,$AM = MB = 40 \, cm$.In $	riangle PMA$,by Pythagoras theorem,$PM^2 + AM^2 = PA^2$.$PM^2 + 40^2 = 50^2 \implies PM^2 = 2500 - 1600 = 900 \implies PM = 30 \, cm$.Since $P$ does not lie between the chords,the distance between the chords is $PN - PM = 10 \, cm$.$PN = 10 + PM = 10 + 30 = 40 \, cm$.In $	riangle PNC$,$PN^2 + NC^2 = PC^2$.$40^2 + NC^2 = 50^2 \implies 1600 + NC^2 = 2500 \implies NC^2 = 900 \implies NC = 30 \, cm$.Since $N$ is the midpoint of $CD$,$CD = 2 	imes NC = 2 	imes 30 = 60 \, cm$.

The radius of a circle with centre $P$ is $50 \, cm$. Centre $P$ is not lying between two parallel chords $AB$ and $CD$. If $AB = 80 \, cm$ and the distance between $AB$ and $CD$ is $10 \, cm$,then find the length of $CD$. (Given $AB > CD$) (in $, cm$)

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