Question

Two chords AB and CD of lengths 8 and 12 respectively of a circle are parallel to each other and are on opposite sides of its centre O. If the distance between the chords AB and CD is 5 then find the radius of the circle.

Answer 1

Given that :- The two chords AB and CD are $8cm$ and $12cm$respectively,

As we have to find the radius of the circle,

Here, first join OA and OC,

Let, r be the radius and O be the centre

Then, draw $OP$ ⊥ $AB$ and $OQ$ ⊥ $CD$

So, the points $P, Q, R$ are collinear,

Therefore,

$PQ=5cm$

Suppose, $OP = x$

$OQ=(5-x)cm$

Also, $AP=PB=4cm$ and $CQ=QD=6cm$

Now, in the right angled triangles $QAP$ and $OCQ$

$OA^{2} =OP^{2} +AP^{2}$ and $OC^{2} =OQ^{2} +CQ^{2}$

∴ $r^{2} =x^{2} +(4)^{2} ...(1)$

and $x^{2} +(5-x)^{2} +(6)^{2} ...(2)$

⇒ $x^{2} +(4)^{2} =(5-x)^{2} +(6)^{2}$

⇒ $x^{2} +16=(5-x)^{2} +36$

⇒ $x=\frac{9}{2}$

Now, by putting $x$ as $\frac{9}{2}$ in equation (1) we get,

⇒ $r^{2} =(\frac{9}{2} )^{2} +(4)^{2}$

⇒ $r^{2} =(20.25)+(16)$

⇒ $r^{2} =36.25$

⇒ $r=6.02$ cm

Thus, the radius of the circle is $6.02 cm$