# Q: 2     Two chords AB and CD of lengths $\small 5\hspace {1mm}cm$ and $\small 11\hspace {1mm}cm$ 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 $\small 6\hspace {1mm}cm$, find the radius of the circle.

M mansi

Given : AB = 5 cm, CD = 11 cm and AB || CD.

Construction: Draw $OM \perp CD \, \, and \, \, \, ON\perp AB$

Proof :

Proof: CD is a chord of circle  and  $OM \perp CD$

Thus, CM = MD = 5.5 cm    (perpendicular from centre bisects chord)

and      AN = NB  = 2.5 cm

Let OM be x.

So, ON = 6 - x              (MN = 6 cm )

In $\triangle$ OCM , using pythagoras,

$OC ^2=CM^2+OM^2$.............................1

and

In $\triangle$ OAN , using pythagoras,

$OA ^2=AN^2+ON^2$.............................2

From 1 and 2,

$CM ^2+OM^2=AN^2+ON^2$            (OC=OA =radii)

$5.5 ^2+x^2=2.5^2+(6-x)^2$

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

$\Rightarrow 30.25-42.25=-12x$

$\Rightarrow -12=-12x$

$\Rightarrow x=1$

From 2, we get

$OC^2=5.5^2+1^2=30.25+1=31.25$

$\Rightarrow OC=\frac{5}{2}\sqrt{5} cm$

OA = OC

Thus, the radius of the circle is $\frac{5}{2}\sqrt{5} cm$

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