Question

In the given figure, $O$ is the centre of the circle. $AB$ and $CD$ are two chords of the circle. $OM$ is perpendicular to $AB$ and $ON$ is perpendicular to $CD$. $AB=24cm,OM=5cm,ON=12cm$. Find the:
(i) radius of the circle.
(ii) length of chord $CD$.

Answer 1

This question is solely based on the property that perpendicular dropped from the centre of the circle on any chord of that circle will bisect the chord.

Solving $\left(i\right)$

$\Rightarrow AM=BM=\frac{AB}{2}=12cm$

Now by applying Pythagorous theorem in $△OMB$

$\Rightarrow O{B}^2=O{M}^2+M{B}^2$

$\Rightarrow R^2=5^2+{12}^2=25+144=169[\because OB=Radius]$

$\Rightarrow R=13$

Solving $\left(ii\right)$

Applying Pythagorous theorem in $△OCN$

$\Rightarrow O{C}^2=O{N}^2+N{C}^2[\because OC=Radius=13]$

$\Rightarrow {13}^2={12}^2+N{C}^2$

$\Rightarrow N{C}^2=169-144=25$

$\Rightarrow NC=5$

Now,since ON is perpendicular to the chord CD

$\Rightarrow$ N is mid point of chord CD

$\Rightarrow CD=2\times NC=2\times 5=10$

Hence, answer of $\left(i\right)$ is $13$ and that of $\left(ii\right)$ is $10$