Question

$AD$ is a diameter of a circle and $AB$ is a chord. If $AD=34cm,AB=30cm$, the distance of $AB$ from the centre of the circle is

• A. $17cm$
• B. $15cm$
• C. $4cm$
• D. $8cm$

Answer 1

The correct option is D $8cm$

It is given that, Diameter, $AD=34$ cm and chord, $AB=30$ cm

Draw a perpendicular $ON$ from $O$ to $AB$. It meets $AB$ at $N$.

$\therefore$ Radius, $AO=\frac{1}{2}AD=\frac{1}{2}\times 34$ cm $=17$ cm.

$\because ON\perp AB$

$\therefore ON$ bisects $AB$ at $N$ and $\angle ANO={90}^0$

So, $AN=\frac{1}{2}AB=\frac{1}{2}\times 30$ cm $=15$ cm

Now, $AO=17$ cm, $AN=15$ cm and $\angle ANO={90}^0$

$\therefore \Delta AON$ is a right one with hypotenuse $AO.$

So, by pythagoras theorem, we have

${AO}^2-{AN}^2={ON}^2\Rightarrow {ON}^2={(17}^2{-15}^2){cm}^2$

$\Rightarrow ON=8$ cm

So, $AB$ is at a distance of $8$ cm from the centre $O$.