CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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 $$CD$$
584887_0e883d8d1f554ab6a0f01ada770592f8.png


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 $$CD$$.
$$\therefore CD = 2\ CM = 2\times 5 \ cm= 10\ cm$$.

615696_584887_ans.png

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image