ICSE Class X Math 2014 Paper

In the figure given below, $O$ is the centre of the circle. $A B$ and $C D$ are two chords of the circle. $O M$ is perpendicular to $A B$ and $O N$ is perpendicular to $C D$ .

$A B=24\text{ cm}, O M=5\text{ cm}, O N=12\text{ cm}. \;\text{ Find }\;$ the :

Question : 50 of 52

Marks: +1, -0

length of chord CD.

Solution:

Now from

\triangle \mathrm{CNO}

;

\mathrm{CO}^2 = \mathrm{ON}^2 + \mathrm{CN}^2

r^2 = (12)^2 + \mathrm{CN}^2

\because \mathrm{AO} = \mathrm{CO} = r

(13)^2 - (12)^2 = \mathrm{CN}^2

169 - 144 = \mathrm{CN}^2

\Rightarrow \quad \mathrm{CN}^2 = 25

\Rightarrow \quad \mathrm{CN} = 5

\mathrm{ON} \perp \mathrm{CD}, \mathrm{N}

is mid point of

\mathrm{CD}

\therefore \mathrm{CD} = 2 \mathrm{CN} = 2 \times 5 = 10 \mathrm{cm}

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