Question

$O$ is the centre of the circle. $AB$ is a chord of the circle. $OM\perp AB$. If $AB=20$ cm, $OM=2\sqrt{11}$ cm, then radius of the circle is

• A. $15$ cm
• B. $12$ cm
• C. $10$ cm
• D. $11$ cm

Answer 1

Answer: (B)

$12$ cm

OA is the radius of the circle

Perpendicular drawn from centre to a chord bisects the chord.

Therefore $AM=\frac{1}{2}AB$

$AM=10cm$

Thus, triangle $OMA$ is right angled triangled at $\angle OMA$.

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

By Pythagoras theorem,

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

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

$OA=\sqrt{{\left(2\sqrt{11}\right)}^2+{10}^2}$

$OA=\sqrt{144}cm$

$OA=12cm$

Radius of a circle is $12cm$.

Option B will be the answer.