Question

In fig, chord $AB\parallel CD$ of a circle with centre $O$ and radius $5$ cm such that $AB=6$cm and $CD=8$cm. if $OP$ $\perp$ AB, $OQ$.$\perp$ $CD$, then $PQ$ in cm is

• A. $4$ cm
• B. $7$ cm
• C. $1$ cm
• D. $3$ cm

Answer 1

The correct option is C $1$ cm

Given- $AB=6$ cm and $CD=8$ cm are the chords of a circle of radius $5$ cm with centre at $O$.

$OP\perp AB$ at $M$ and $OQ\perp CD$ at $N$.

To find out - the length of $PQ=?$

Solution-

We join $OC$ and $OA$.

So, $OC=OA=5$ cm, since $OC$ and $OA$ are radii.

$\Delta OAP$ and $\Delta OCQ$ are right ones, since $OP\perp AB$ at $P$ and $OQ\perp CD$ at $Q$.

Now $AP=\frac{1}{2}AB=\frac{1}{2}\times 6$ cm $=3$ cm and $CQ=\frac{1}{2}CD=\frac{1}{2}\times 8$ cm $=4$ cm, since the perpendicular from the centre of a circle to a chord bisects the latter.

So, in $\Delta OAP$, by Pythagoras theorem, we have

$OP=\sqrt{{OA}^2-{AP}^2}=\sqrt{5^2-4^2}$ cm $=3$ cm

Again in $\Delta OCQ$, by Pythagoras theorem, we have

$OQ=\sqrt{{OC}^2-{CQ}^2}=\sqrt{5^2-3^2}$ cm $=4$ cm.

$\therefore PQ=OP-OQ=(4-3)$ cm $=1$ cm