ICSE Class X Math 2014 Solved 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 $C D$ .

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

Question : 50 of 52

Marks: +1, -0

length of chord CD.

Solution:

Now from

∆ {\CNO}

;

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

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

\;( ∵ {\AO}= {\CO}=r)

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

169-144=\; {\CN}^2

⇒ \;\; {\CN}^2=\;25

⇒ \;\; {\CN}=\;5

{\ON} \⟂ {\CD}, {\N}

is mid point of

{\CD}

∴ {\CD}=2 {\CN}=2 × 5=10 {\cm}

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