Question

AB and CD are two parallel chords of a circle which are on either sides of the centre. Such that AB = $10$ cm and CD = $24$ cm. Find the radius if the distance between AB and CD is $17$ cm.

Answer 1

Let $OP\perp AB$ and $OQ\perp CD$

CQ = QD = $12$ cm

AP = BP = $5$ cm

Given PQ = $17$ , hence OP + OQ = $17$

In $△AOP$

$A{O}^2=O{P}^2+A{P}^2$

$r^2=O{P}^2+25-eq.1$

In $△COQ$

$C{O}^2=O{Q}^2+C{Q}^2$

$r^2=(17–OP{)}^2+144-eq.2$

$Eq.2–eq.1$

${17}^2+O{P}^2-2OP\cdot 17-O{P}^2+144-25=0$

$17(17–2OP)+119=0$

$⟹289-34OP+119\Rightarrow 34OP=408$

$⟹OP=12$

Substitute in $e.q.1$

$⟹r^2=144+25=169$

$r=13$