Question

A chord of a circle is $12$ cm in length and its distance from the centre is $8$ cm. The length of the chord of the same circle which is at a distance of $6$ cm from the centre is

• A. $30$ cm
• B. $24$ cm
• C. $16$ cm
• D. $18$ cm

Answer 1

The correct option is C $16$ cm

OA is the radius of a circle.

Perpendicular drawn from centre bisects the chord.

Therefore $AM=\frac{1}{2}AB=\frac{12}{2}=6cm$

$AM=6cm$

And, Triangle $OMA$ is right angled at M$

$\angle OMA={90}^{\circ }$

By Pythagoras theorem,

${OA}^2={OM}^2+{MA}^2$

$OA=\sqrt{{OM}^2+{MA}^2}$

$OA=\sqrt{8^2+6^2}$

$OA=\sqrt{100}cm$

$OA=10cm$

Radius of a circle is $10cm$.

Therefore $OD=10cm$

Similarly $OP$ is perpendicular to $CD$ so it bisects $CD$.

(Since, Perpendicular drawn from centre bisects the chord.)

Triangle $OPD$ is right angled at $P$

$\angle OPD={90}^{\circ }$

By Pythagoras theorem,

${OD}^2={OP}^2+{PD}^2$

$PD=\sqrt{{OD}^2-{OP}^2}$

$PD=\sqrt{{10}^2-6^2}$

$PD=\sqrt{64}cm$

$PD=8cm$

$CD=2.PD=16cm$.

Therefore length of chord $CD=16cm$.

So, option C will be the answer.