Question

$AB$ is chord of a circle with centre $O$ and radius $17$ cm. If $OM\perp AB$ and $OM=8$ cm, then the length of chord $AB$ is

• A. $12$ cm
• B. $30$ cm
• C. $15$ cm
• D. $24$ cm

Answer 1

Answer: (B)

$30$ cm

Given- $O$ is the centre of a circle of radius $17$ cm.

$OM\perp AB$ when $AB$ is a chord of the given circle.

To find out- the length of $AB=?$

Solution-

We join $OA$.

So, $OA$ is a radius of the given circle.

$\therefore OA=17$ cm

Now $OM\perp AB$

$\Rightarrow \angle OMA={90}^o$

Therefore, $\Delta OMA$ is a right one with $OA$ as hypotenuse.

Therefore by Pythagoras theorem, we get

$AM=\sqrt{{OA}^2-{AM}^2}=\sqrt{{17}^2-8^2}cm$

$=15cm$ .

Again $M$ is the mid point of $AB$, since $OM\perp AB$ and we

know that the perpendicular, from the centre of a circle to any of its chords, bisects the latter.

$\therefore AB=2AM=2\times 15$ cm $=30$ cm