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  • Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Q 2. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between $A B$ and $C D$ is 6 cm , find the radius of the circle.

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Given: $A B=5 \mathrm{~cm}, C D=11 \mathrm{~cm}$ and $A B \| C D$.
To find Radius (OA).
Construction: Draw $O M \perp C D$ and $O N \perp A B$
Proof :

          

Proof: CD is a chord of the circle and $O M \perp C D$
Thus, $\mathrm{CM}=\mathrm{MD}=5.5 \mathrm{~cm}$ (perpendicular from centre bisects chord)
and $\mathrm{AN}=\mathrm{NB}=2.5 \mathrm{~cm}$
Let $O M$ be x .
So, $\mathrm{ON}=6-\mathrm{x} \quad(\mathrm{MN}=6 \mathrm{~cm})$
In $\triangle$ OCM, using Pythagoras,
$O C^2=C M^2+O M^2$-----------1

and

In $\triangle$ OAN, using Pythagoras,
$O A^2=A N^2+O N^2$------(2)
From 1 and 2,
$C M^2+O M^2=A N^2+O N^2$ (OC=OA= radii)
$5.5^2+x^2=2.5^2+(6-x)^2$
$\Rightarrow 30.25+x^2=6.25+36+x^2-12 x$
$\Rightarrow 30.25-42.25=-12 x$
$\Rightarrow-12=-12 x$

$\Rightarrow x=1$
From 2, we get
$O C^2=5.5^2+1^2=30.25+1=31.25$
$\Rightarrow O C=\frac{5}{2} \sqrt{5} \mathrm{~cm}$
$\mathrm{OA}=\mathrm{OC}$
Thus, the radius of the circle is $\frac{5}{2} \sqrt{5} \mathrm{~cm}$.

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