Question

The radius of a circle with centre $O$ is $41$. The length of the chord $PQ$ of same circle is $80$. The distance of chord $PQ$ from centre $O$ is

• A. $81$ cm
• B. $27$ cm
• C. $3$ cm
• D. $9$ cm

Answer 1

The correct option is D $9$ cm

Given- $PQ=80$ cm is a chord of the circle with centre $O$ and radius $=r=41$ cm.

$ON$ is the perpendicular distance of $PQ$ from $O$

To find out- $ON=?$

Solution:

We know that a chord of a circle is bisected by the line which is the perpendicular distance of the chord from the centre of the given circle.

$\therefore PN=QN=\frac{80}{2}$ cm $=40$ cm.

Now in $\Delta OPN$, we have

$\angle OPN=1$ right angle i.e. $\Delta OPN$ is a right one with hypotenuse as $PO$.

$\therefore$ applying Pythagoras theorem, we have

$ON=\sqrt{{OP}^2-{PN}^2}=\sqrt{{41}^2-{4O}^2}$ cm $=9$ cm