AB and CD are two parallel chords of a circle | vedclass

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(A) Let $P$ be the centre of the circle with radius $r = 25\,cm$.Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively.Since the perpe

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(A) Let $P$ be the centre of the circle with radius $r = 25\,cm$.Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively.Since the perpendicular from the centre to a chord bisects the chord,$AM = MB = AB/2 = 40/2 = 20\,cm$ and $CN = ND = CD/2 = 30/2 = 15\,cm$.In right-angled $	riangle PMA$,by Pythagoras theorem: $PM^2 + AM^2 = PA^2 \implies PM^2 + 20^2 = 25^2 \implies PM^2 = 625 - 400 = 225 \implies PM = 15\,cm$.In right-angled $	riangle PNC$,by Pythagoras theorem: $PN^2 + CN^2 = PC^2 \implies PN^2 + 15^2 = 25^2 \implies PN^2 = 625 - 225 = 400 \implies PN = 20\,cm$.Since the centre $P$ is not between the chords,the distance between the chords is $|PN - PM| = |20 - 15| = 5\,cm$.

$AB$ and $CD$ are two parallel chords of a circle with centre $P$. Centre $P$ is not between the two chords $AB$ and $CD$. If $AB = 40\,cm$,$CD = 30\,cm$ and the radius of the circle is $25\,cm$,find the distance between $AB$ and $CD$. (in $,cm$)

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