Two chords AB and CD of lengths 5, cm and 11, | vedclass

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Question

(B) Let the circle have centre $O$. Chords $AB$ and $CD$ are parallel,and the distance between them is $6\, cm$. Given $AB = 5\, cm$ and $CD = 11\, cm

Vedclass · Accepted Answer

$\frac{5 \sqrt{5}}{2}\, cm$

Vedclass · Answer

(B) Let the circle have centre $O$. Chords $AB$ and $CD$ are parallel,and the distance between them is $6\, cm$. Given $AB = 5\, cm$ and $CD = 11\, cm$.Let $r$ be the radius of the circle.Draw $OP \perp AB$ and $OQ \perp CD$. Since the chords are on opposite sides of the centre,$P, O, Q$ are collinear and $PQ = 6\, cm$.Let $OQ = x\, cm$. Then $OP = (6 - x)\, cm$.The perpendicular from the centre to a chord bisects the chord.So,$AP = \frac{1}{2} AB = \frac{5}{2}\, cm$ and $CQ = \frac{1}{2} CD = \frac{11}{2}\, cm$.In right $\Delta CQO$,$r^2 = CQ^2 + OQ^2 = (\frac{11}{2})^2 + x^2 = \frac{121}{4} + x^2$  --- $(1)$In right $\Delta APO$,$r^2 = AP^2 + OP^2 = (\frac{5}{2})^2 + (6 - x)^2 = \frac{25}{4} + 36 - 12x + x^2$ --- $(2)$Equating $(1)$ and $(2)$:$\frac{121}{4} + x^2 = \frac{25}{4} + 36 - 12x + x^2$$\frac{121}{4} - \frac{25}{4} - 36 = -12x$$\frac{96}{4} - 36 = -12x \Rightarrow 24 - 36 = -12x \Rightarrow -12 = -12x \Rightarrow x = 1\, cm$.Substituting $x = 1$ in $(1)$:$r^2 = \frac{121}{4} + 1^2 = \frac{121 + 4}{4} = \frac{125}{4}$.$r = \sqrt{\frac{125}{4}} = \frac{5\sqrt{5}}{2}\, cm$.

Two chords $AB$ and $CD$ of lengths $5\, cm$ and $11\, cm$ respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between $AB$ and $CD$ is $6\, cm$,find the radius of the circle.

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