Question

In a circle of radius $25$ cm two parallel chords of the length $14$ cm and $48$ cm respectively, are drawn on the same side of the centre. The distance between them is

• A. $14$ cm
• B. $24$ cm
• C. $17$ cm
• D. $31$ cm

Answer 1

The correct option is C $17$ cm

Given- $AB=14$ cm and $CD=48$ cm are the chords of a circle of radius $25$ cm with centre at $O$.

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

To find out -

If the length of $PQ=?$

Solution-

We join $OC$ and $OA$.

So, $OC=OA=25$ 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 $AP=\frac{1}{2}AB=\frac{1}{2}\times 14$ cm $=7$ cm and

$CQ=\frac{1}{2}CD=\frac{1}{2}\times 48$ cm $=24$ 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

$OP=\sqrt{{OA}^2-{AP}^2}=\sqrt{{25}^2-7^2}$ cm $=24$ cm

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

$OQ=\sqrt{{OC}^2-{CQ}^2}=\sqrt{{25}^2-{24}^2}$ cm $=7$ cm

$\therefore PQ=OP-OQ=(24-7)$ cm $=17$ cm