# CAT 2018 Set-1 | Question: 73

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In a circle, two parallel chords on the same side of a diameter have lengths $4$ cm and $6$ cm. If the distance between these chords is $1$ cm, then the radius of the circle, in cm, is

1. $\sqrt{12}$
2. $\sqrt{14}$
3. $\sqrt{13}$
4. $\sqrt{11}$

Given that, the length of long chord is $6 \; \text{cm},$  the length of small chord is $4 \; \text{cm},$ and the distance between these chords is $1\;\text{cm}.$

Let the radius of the circle be $r\text{’} \; \text{cm},$ and let the distance of long chord from the center of circle be $k\text{’} \; \text{cm}.$

We can draw the diagram,

We know that the perpendicular from the center of the circle to the chords bisects the chords.

In $\triangle \text{OFB},$ apply the Pythagoras’ theorem :

$\text{(OB)}^{2} = \text{(OF)}^{2} + \text{(FB)}^{2}$

$\Rightarrow r^{2} = (k+1)^{2} + 2^{2}$

$\Rightarrow r^{2} = k^{2} + 1 + 2k + 4$

$\Rightarrow r^{2} = k^{2} + 2k + 5 \quad \longrightarrow (1)$

In $\triangle \text{OED},$ apply the Pythagoras’ theorem :

$\text{(OD)}^{2} = \text{(OE)}^{2} + \text{(ED)}^{2}$

$\Rightarrow r^{2} = k^{2} + 3^{3}$

$\Rightarrow r^{2} = k^{2} + 9 \quad \longrightarrow (2)$

On equating the equation $(1),$ and $(2),$ we get,

$k^{2} + 2k + 5 = k^{2} + 9$

$\Rightarrow 2k = 4$

$\Rightarrow \boxed {k= 2 \;\text{cm}}$

Put the value of $k$ in equation $(2),$ we get,

$r^{2} = 2^{2} + 9$

$\Rightarrow r^{2} = 4 + 9$

$\Rightarrow r^{2} = 13$

$\Rightarrow \boxed { r = \sqrt{13}\; \text{cm}}$

$\therefore$ The radius of the circle is $\sqrt{13}\;\text{cm}.$

Correct Answer $: \text{C}$

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