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

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.

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M mansi

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

To find Radius (OA).

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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