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(C) Let the circle have centre $O$. Let the parallel chords be $AB = 6\, cm$ and $CD = 8\, cm$.Draw $OP \perp AB$ and $OQ \perp CD$.Since the perpendi

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$3$

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(C) Let the circle have centre $O$. Let the parallel chords be $AB = 6\, cm$ and $CD = 8\, cm$.Draw $OP \perp AB$ and $OQ \perp CD$.Since the perpendicular from the centre to a chord bisects the chord:$AP = \frac{1}{2} AB = \frac{1}{2}(6\, cm) = 3\, cm$$CQ = \frac{1}{2} CD = \frac{1}{2}(8\, cm) = 4\, cm$Given that the smaller chord is at a distance of $4\, cm$ from the centre,$OP = 4\, cm$.In right-angled $\Delta OPA$,by Pythagoras theorem:$OA^2 = OP^2 + AP^2$$r^2 = 4^2 + 3^2 = 16 + 9 = 25$$r = 5\, cm$ (where $r$ is the radius of the circle).Now,in right-angled $\Delta OQC$:$OC^2 = OQ^2 + CQ^2$$r^2 = OQ^2 + 4^2$$5^2 = OQ^2 + 16$$25 = OQ^2 + 16$$OQ^2 = 25 - 16 = 9$$OQ = 3\, cm$.Thus,the distance of the other chord from the centre is $3\, cm$.

The lengths of two parallel chords of a circle are $6\, cm$ and $8\, cm$. If the smaller chord is at a distance of $4\, cm$ from the centre,what is the distance of the other chord from the centre (in $, cm$)?

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