Question

PQ and RS are two parallel chords of a circle whose centre is O and radius is $10$ cm. If PQ $=16$ cm and RS $=12$ cm. Then the distance between PQ and RS, if they lie.
(i) on the same side of the centre O and
(ii) on the opposite of the centre O are respectively

• A. $8$ cm & $14$ cm
• B. $4$ cm & $14$ cm
• C. $2$ cm & $14$ cm
• D. $2$ cm & $28$ cm

Answer 1

The correct option is C $2$ cm & $14$ cm

We join $OQ$ & $OS$, drop perpendicular from O to $PQ$ & $RS$.

The perpendiculars meet $PQ$ & $RS$ at M & N respectively.

Since OM & ON are perpendiculars to $PQ$ & $RS$ who are

parallel lines, M, N & O will be on the same straight line

and disance between $PQ$ & $RS$ is $MN$.........(i) and $\angle ONQ={90}^o=\angle OMQ$......(ii)

Again $M$ & $N$ are mid points of $PQ$ & $RS$ respectively since $OM\perp PQ$ & $ON\perp RS$

respectively and the perpendicular, dropped from the center of a circle to any of its chord,

bisects the latter.

So $QM=\frac{1}{2}$PQ$=\frac{1}{2}\times 16$ cm $=8$ cm and $SN=\frac{1}{2}RS=\frac{1}{2}\times 12$ cm$=6$ cm.

$\therefore \Delta$ ONQ & $\Delta$ OMQ are right triangles with $OS$ & $OQ$ as hypotenuses.(from ii)

So, by Pythagoras theorem, we get $ON=\sqrt{{OS}^2-{SN}^2}=\sqrt{{10}^2-6^2}$ cm $=8$ cm and $OM=\sqrt{{OQ}^2-{QM}^2}=\sqrt{{10}^2-8^2}$ cm $=6$ cm.

Now two cases arise- (i) $PQ$ & $RS$ are to the opposite side of the centre O.(fig I)

Here $MN=OM+ON$=(6+8)$cm$=14$ cm (from i) or

(ii) $PQ$ & $RS$ are to the same side of the centre O. (fig II)

Here $MN=ON-OM=(8-6)$ cm$=2$ cm.

So the distance between $PQ$ & $RS$ $=14$ cm when $PQ$ & $RS$ are to the opposite side of the centre O

and the distance between $PQ$ & $RS$ $=2$ cm when $PQ$ & $RS$ are to the same side of the centre O.

Ans- Option C.