Question

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

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

Answer 1

Answer: (B)

$30$ cm

Radius of Circle is $17cm$.

Therefore $OA=17cm$.

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

Therefore $AM=MB$.

Triangle OMA is right angled at M.

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

By Pythagoras theorem,

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

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

$MA=\sqrt{{17}^2-8^2}$

$MA=\sqrt{225}$

$MA=15$

Therefore the length of chord $AB=2.MA$.

$AB=30cm$.

option B will be the answer.