Question

A chord $AB$ is at a distance of $6$ cm from the centre of a circle whose radius is $6$ cm less than that of the chord $AB$. Then the length of the chord $AB$ is

• A. $8$ cm
• B. $32$ cm
• C. $24$ cm
• D. $16$ cm

Answer 1

The correct option is D $16$ cm

Given- $AB$ is a chord of a circle with centre $O$.
$ON$ is the perpendicular from $O$ to $AB$ at $N$.

$ON=6$ cm, $AB=(OA+6)$ cm,$ON\perp AB$

To find out- the length of the radius of the circle $=?$

Solution-

Let $OA=x-6$ cm i.e. $AB=x$ cm.

$\therefore AN=\frac{1}{2}AB=\frac{x}{2}$ cm, since the perpendicular from the centre of a circle to a chord bisects the lattar.

Now in $\Delta OAN$, we have

$\angle ANO={90}^o$ as $ON\perp AB$.

So, $\Delta OAN$ is a right one with $OA$ as hypotenuse.

$\therefore$ applying pythagoras theorem, we have

$OA=$ radius of the given circle $=\sqrt{{ON}^2+{AN}^2}=\sqrt{6^2+{AN}^2}$

$\Rightarrow x-6=\sqrt{6^2+{\left(\frac{x}{2}\right)}^2}$

$\Rightarrow x(3x-48)=0$

$\Rightarrow x=(0,16)$ cm

We reject $x=0$, since it is\quad a finite length.

So, $x=AB=16$ cm.