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

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Question

(C) Let $P$ be the centre of the circle and $r = 26\,cm$ be the radius.Draw perpendiculars $PM$ and $PN$ from $P$ to chords $AB$ and $CD$ respectively

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(C) Let $P$ be the centre of the circle and $r = 26\,cm$ be the radius.Draw perpendiculars $PM$ and $PN$ from $P$ to chords $AB$ and $CD$ respectively.Since the perpendicular from the centre to a chord bisects the chord,we have:$AM = MB = \frac{AB}{2} = \frac{20}{2} = 10\,cm$$CN = ND = \frac{CD}{2} = \frac{48}{2} = 24\,cm$In right-angled $	riangle PMA$,by Pythagoras theorem:$PM^2 + AM^2 = PA^2$$PM^2 + 10^2 = 26^2$$PM^2 = 676 - 100 = 576$$PM = \sqrt{576} = 24\,cm$In right-angled $	riangle PNC$,by Pythagoras theorem:$PN^2 + CN^2 = PC^2$$PN^2 + 24^2 = 26^2$$PN^2 = 676 - 576 = 100$$PN = \sqrt{100} = 10\,cm$Since the centre $P$ lies between the chords,the distance between $AB$ and $CD$ is $PM + PN = 24 + 10 = 34\,cm$.

$AB$ and $CD$ are two parallel chords of a circle with centre $P$. Also,the centre $P$ lies between the chords $AB$ and $CD$. If $AB = 20\,cm$,$CD = 48\,cm$,and the radius of the circle is $26\,cm$,find the distance between $AB$ and $CD$.

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