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

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(C) Let the radius of the circle be $r$ and the centre be $O$.Let $OM \perp AB$ and $ON \perp CD$,where $M$ and $N$ are the midpoints of chords $AB$ a

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$13$

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(C) Let the radius of the circle be $r$ and the centre be $O$.Let $OM \perp AB$ and $ON \perp CD$,where $M$ and $N$ are the midpoints of chords $AB$ and $CD$ respectively.Given $AB = 10 \, cm$,so $AM = MB = 5 \, cm$.Given $CD = 24 \, cm$,so $CN = ND = 12 \, cm$.Let $ON = x$. Since the chords are on opposite sides of the centre,$OM = 17 - x$.In right-angled $\Delta ONA$,$OA^2 = ON^2 + CN^2 \implies r^2 = x^2 + 12^2 = x^2 + 144 \dots(1)$.In right-angled $\Delta OMA$,$OA^2 = OM^2 + AM^2 \implies r^2 = (17 - x)^2 + 5^2 = 289 + x^2 - 34x + 25 = x^2 - 34x + 314 \dots(2)$.Equating $(1)$ and $(2)$:$x^2 + 144 = x^2 - 34x + 314$$34x = 314 - 144$$34x = 170$$x = 5 \, cm$.Substituting $x = 5$ in $(1)$:$r^2 = 5^2 + 144 = 25 + 144 = 169$$r = \sqrt{169} = 13 \, cm$.

$AB$ and $CD$ are two parallel chords of a circle lying on the opposite side of the centre and the distance between them is $17 \, cm$. The lengths of $AB$ and $CD$ are $10 \, cm$ and $24 \, cm$ respectively. The radius (in $cm$) of the circle is

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