Two chords $AB$ and $CD$ of lengths $5cm$ and $11cm$ 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 $6cm$, find the radius of the circle.

Join $OA$ and $OC$.

Let the radius of the circle be $r$ cm and $O$ be the centre 

Draw $OP\perp AB$ and $OQ\perp CD$. 

We know, $OQ\perp CD$, $OP\perp AB$ and $AB\parallel CD$. 

Therefore, points $P,O$ and $Q$ are collinear. So, $PQ=6$ cm. 

Let $OP=x$. 

Then, $OQ=(6–x)$ cm. 
And $OA=OC=r$. 

Also, $AP=PB=2.5$ cm and $CQ=QD=5.5$ cm.

(Perpendicular from the centre to a chord of the circle bisects the chord.)

  
In right triangles $QAP$ and $OCQ$, we have

$O{A}^2=O{P}^2+A{P}^2$ and $O{C}^2=O{Q}^2+C{Q}^2$

$\therefore r^2=x^2+(2.5{)}^2$                 ..... (1) 

and $r^2=(6-x{)}^2+(5.5{)}^2$      ..... (2) 

$\Rightarrow x^2+(2.5{)}^2=(6-x{)}^2+(5.5{)}^2$

$\Rightarrow x^2+6.25=36-12x+x^2+30.25$

$12x=60$

$\therefore x=5$

Putting $x=5$ in (1), we get 

$r^2=52+(2.5{)}^2=25+6.25=31.25$

$\Rightarrow r^2=31.25\Rightarrow r=5.6$
Hence, the radius of the circle is $5.6$ cm