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The length of two parallel chords of a circle are $6\ cm$ and $8\ cm$. If the smaller chord is at a distance of $4\ cm$ from the centre, what is the distance of the other chord from the centre?
Given:
The length of two parallel chords of a circle are $6\ cm$ and $8\ cm$.
The smaller chord is at a distance of $4\ cm$ from the centre.
To do:
We have to find the distance of the other chord from the centre.
Solution:
Let a circle with centre $O$ and two parallel chords $AB$ and $CD$ are $AB = 6\ cm, CD = 8\ cm$
Let $OL \perp AB$ and $OM \perp CD$
Therefore,
$OL = 4\ cm$
Let $OM = x\ cm$ and $r$ be the radius of the circle.
In right angled triangle $\mathrm{OAL}$,
$\mathrm{OA}^{2}=\mathrm{OL}^{2}+\mathrm{AL}^{2}$
$=4^{2}+(\frac{6}{2})^{2}$
$=16+9$
$=25$.......(i)
In right angled triangle $\Delta \mathrm{OMC}$,
$\mathrm{OC}^{2}=\mathrm{OM}^{2}+\mathrm{CM}^{2}$
$r^{2}=x^{2}+(\frac{8}{2})^{2}$
$=x^{2}+(4)^{2}$
$=x^{2}+16$..........(ii)
From (i) and (ii), we get,
$x^{2}+16=25$
$\Rightarrow x^{2}=25-16=9$
$\Rightarrow x^{2}=(3)^{2}$
$x=3 \mathrm{~cm}$
Distance $=3 \mathrm{~cm}$
The distance of the other chord from the centre is $3\ cm$.