Question

# In the figure given, $$O$$ is the centre of the circle. $$AB$$ and $$CD$$ are two chords of the circle. $$OM$$ is perpendicular to $$CD$$ and $$ON$$ is perpendicular to $$AB$$. $$AB = 24\ cm, ON = 5\ cm, OM = 12\ cm$$. Find the length of chord $$CD$$

Solution

## Given: $$AB = 24\ cm; ON = 5\ cm, OM = 12\ cm$$.$$\because ON \perp AB$$$$N$$ is the mid point of $$AB$$.$$\therefore AN =\dfrac{24}{2}cm= 12\ cm$$.Now from $$\triangle ANO,$$$$AO^{2} = ON^{2} + AN^{2}$$$$\Rightarrow r^{2} = 5^{2} + 12^{2}$$      $$(\because AO = CO = r)$$$$\Rightarrow r^2 =25+144$$$$\Rightarrow r=13$$So,$$AO=CO=13\ cm$$From $$\Delta CMO$$,$$CM^2=CO^2-OM^2$$$$\Rightarrow CM^2=13^2 - 12^2$$$$\Rightarrow CM^2=169-144$$$$\Rightarrow CM^{2} = 25$$$$\Rightarrow CM = 5$$As $$OM\perp CD, M$$ is the mid point of $$CD$$.$$\therefore CD = 2\ CM = 2\times 5 \ cm= 10\ cm$$.Mathematics

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