Question

$O$ is the centre of the circle having radius $5$ cm. $OM$ is a $\perp$ on chord $AB$. If $OM=4$ cm, then the length of the chord $AB$ is equal to

• A. $5$ cm
• B. $6$ cm
• C. $8$ cm
• D. $10$ cm

Answer 1

The correct option is B $6$ cm

Radius of a circle is 5 cm.

Therefore OB = 5 cm.

Perpendicular drawn to the chord from the centre bisects the chord.

Therefore AM = MB

Triangle OMB is right angled triangle at M

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

By Pythagoras theorem;

${OB}^2={OM}^2+{MB}^2$

$MB=\sqrt{{OB}^2-{OM}^2}$

$MB=\sqrt{5^2-4^2}$

$MB=3cm$

So, the length of chord $AB=2.\left(MB\right)=6cm$.

Option B will be the answer.