Question

A chord of length $L$ cm is drawn in a circle of radius $R$ cm. The distance of the chord from the center of the circle is

• A. $\sqrt{L^2}-\frac{R^2}{4}$
• B. $\sqrt{\frac{4R^2-L^2}{4}}$
• C. $\beta$
• D. $\sqrt{R^2}+L^2-RL$

Answer 1

The correct option is B $\sqrt{\frac{4R^2-L^2}{4}}$

Suppose $AB$ is the chord, $O$ is the centre of the circle. If $OM$ is the perpendicular drawn from $O$ to $AB$ as its distance, then $OAM$ and $OMB$ are both right angled triangles.
Since, length of the chord is $L$, therefore $AM=MB=\frac{L}{2}$
If r is the radius of the circle, then
$=>{OB}^2={OM}^2+{MB}^2$
$=>R^2={OM}^2+{\left(\frac{L}{2}\right)}^2$
$=>{OM}^2=R^2-\frac{L^2}{4}$

$=>{OM}^2=\frac{4R^2-L^2}{4}$
=> Distance $=OM=\frac{\sqrt{4R^2-L^2}}{2}$