Question

$AB$ and $CD$ are two parallel chords of a circle such that $AB=10cm$ and $CD=24cm.$ If the chords are on opposite sides of the centre and the distance between them is $17cm,$ find the radius of the circle. $\left[3\text{Marks}\right]$

Answer 1

Let $O$ be the centre of the given circle and let its radius be $rcm.$ Draw $OP\perp AB$ and $OQ\perp CD.$

Since, $OP\perp AB,OQ\perp CD$ and $AB\parallel CD,$ therefore, points $P,O$ and $Q$ are collinear.

Given: $PQ=17cm$

Let $OP=xcm.$ Then, $OQ=17-xcm.$

Join $OA$ and $OC$. Then, $OA=OC=r$.

Since, the perpendicular from the centre to a chord of the circle bisects the chord.

$\Rightarrow AP=PB=5cm$ and $CQ=QD=12cm$. $\left[1\text{Mark}\right]$

In right triangle, $\Delta OAP,$ we have

${OA}^2={OP}^2+{AP}^2$

In right triangle, $\Delta OCQ,$ we have

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

$\Rightarrow r^2=x^2+5^2$ … (i)

and, $r^2={(17-x)}^2+{12}^2$ ... (ii)

$\Rightarrow x^2+5^2={(17-x)}^2+{12}^2$ [Equating (i) and (ii)] $\left[1\text{Mark}\right]$

$\Rightarrow x^2+25=289-34x+x^2+144$

$\Rightarrow 34x=408\Rightarrow x=\frac{408}{34}=12cm$

Putting $x=12cm$ in equation (i), we get

$r^2=144+25=169$

$\Rightarrow r=13cm$.

Hence, the radius of the circle is $13cm$. $\left[1\text{Mark}\right]$