Q : 3  The lengths of two parallel chords of a circle are \small 6\hspace {1mm}cm and \small 8\hspace {1mm}cm. If the smaller chord is at distance \small 4\hspace {1mm}cm from the centre, what is the distance of the other chord from the centre?

Answers (1)

Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB || CD.

To find: Length of ON

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

Proof : 

          

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

    Thus, CM = MD = 3 cm    (perpendicular from centre bisects chord)

 and      AN = NB  = 4 cm

       Let MN be x.

 So, ON = 4 - x              (MN = 4 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)

      \Rightarrow 3 ^2+4^2=4^2+(4-x)^2

\Rightarrow 9+16=16+16+x^2-8x

\Rightarrow 9=16+x^2-8x

\Rightarrow x^2-8x+7=0

\Rightarrow x^2-7x-x+7=0

\Rightarrow x(x-7)-1(x-7)=0

\Rightarrow (x-1)(x-7)=0

\Rightarrow x=1,7

 So, x=1  (since  x\neq 7> OM)

ON =4-x =4-1=3 cm

Hence, second chord is 3 cm away from centre.

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