# 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?

S seema garhwal

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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