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

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(D) Let the centre of the circle be $P$ and the radius $r = 25 \, cm$. Draw perpendiculars from $P$ to the chords $AB$ and $CD$,meeting them at points

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(D) Let the centre of the circle be $P$ and the radius $r = 25 \, cm$. Draw perpendiculars from $P$ to the chords $AB$ and $CD$,meeting them at points $M$ and $N$ respectively. Since the perpendicular from the centre to a chord bisects the chord,we have: $AM = MB = \frac{AB}{2} = \frac{48}{2} = 24 \, cm$ $CN = ND = \frac{CD}{2} = \frac{40}{2} = 20 \, cm$ In right-angled triangle $	riangle PMA$: $PM^2 + AM^2 = PA^2$ $PM^2 + 24^2 = 25^2$ $PM^2 + 576 = 625$ $PM^2 = 49 \implies PM = 7 \, cm$ In right-angled triangle $	riangle PNC$: $PN^2 + CN^2 = PC^2$ $PN^2 + 20^2 = 25^2$ $PN^2 + 400 = 625$ $PN^2 = 225 \implies PN = 15 \, cm$ Since the centre $P$ lies between the chords,the distance between $AB$ and $CD$ is $PM + PN = 7 \, cm + 15 \, cm = 22 \, cm$.

$AB$ and $CD$ are two parallel chords of a circle with centre $P$. Also,the centre $P$ lies between $AB$ and $CD$. If $AB = 48 \, cm$,$CD = 40 \, cm$,and the radius of the circle is $25 \, cm$,find the distance between $AB$ and $CD$. (in $, cm$)

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