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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Updated on: 10-Oct-2022

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