Two chords $ \mathrm{AB} $ and $ \mathrm{CD} $ of lengths $ 5 \mathrm{~cm} $ and $ 11 \mathrm{~cm} $ respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between $ \mathrm{AB} $ and $ C D $ is $ 6 \mathrm{~cm} $, find the radius of the circle.
Given:
Two chords $AB$ and $CD$ of lengths $5\ cm$ and $11\ cm$ respectively of a circle are parallel to each other and are opposite side of its centre.
The distance between $AB$ and $CD$ is $6\ cm$.
To do:
We have to find the radius of the circle.
Solution:
Let $r$ be the radius of the circle with centre $O$.
Two parallel chords $AB = 5\ cm, CD = 11\ cm$
Let $OL \perp AB$ and $OM \perp CD$
$LM = 6\ cm$
Let $OM = x$
This implies,
$OL = 6 - x$
In right angled triangle $\mathrm{OAL}$,
$\mathrm{OA}^{2}=\mathrm{OL}^{2}+\mathrm{AL}^{2}$
$r^{2}=(6-x)^{2}+(\frac{5}{2})^{2}$
$=36-12 x+x^{2}+\frac{25}{4}$............(i)
Similarly,
In right angled $\Delta \mathrm{OCM}$,
$r^{2}=x^{2}+(\frac{11}{2})^{2}$
$=x^{2}+\frac{121}{4}$..............(ii)
From (i) and (ii), we get,
$x^{2}+\frac{121}{4}=36-12 x+x^{2}+\frac{25}{4}$
$\Rightarrow \frac{121}{4}-\frac{25}{4}-36=-12 x$
$\Rightarrow \frac{96}{4}-\frac{36}{1}=-12 x$
$\Rightarrow 12 x=36-24=12$
$x=\frac{12}{12}=1$
$\Rightarrow r^{2}=\mathrm{CM}^{2}+\mathrm{OM}^{2}$
$=(\frac{11}{2})^{2}+(1)^{2}$
$=\frac{121}{4}+1$
$=\frac{125}{4} \mathrm{~cm}$
$\Rightarrow r=\sqrt{\frac{125}{4}}$
$=\frac{\sqrt{125}}{2}$
$=\frac{\sqrt{25 \times 5}}{2}$
$=\frac{5}{2} \sqrt{5} \mathrm{~cm}$
The radius of the circle is $\frac{5}{2} \sqrt{5} \mathrm{~cm}$.
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