`P Q\ a n d\ R S` are two parallel chords of a circle whose centre is `O` and radius is `10\ c mdot` If `P Q=16\ c m\ a n d\ R S=12\ c m ,` find the distance between `P Q\ a n d\ R S ,` if they lie: on opposite side of the centre `O`

Explanation

We join `OQ \and\ OS`
drop perpendicular from `O to PQ and RS.`
The perpendiculars meet `PQ \and\ RS` at `M \and\ N` respectively.
Since, `OM \and\ ON` are perpendiculars to `PQ and RS`
who are
parallel lines, `M, N \and\ O` will be on the same straight line
disance between `PQ \and\ RS \is\ MN`.........(i)
`/_ONQ=90^@=/_OMQ`......(ii)
Again `M \and\ N` are mid points of `PQ \and\ RS` respectively.
since `OM_|_PQ \and\ ON_|_RS`
respectively and the perpendicular, dropped from the center of a circle to any of its chord, bisects the latter.
So `QM=1/2PQ= 1/2xx16 cm =8 cm`
`SN=1/2 RS=1/2 xx12 cm=6 cm.`
`/_\ONQ \and\ /_\OMQ` are right triangles with `OS \and\ OQ` as hypotenuses.(from ii)

So, by Pythagoras theorem, we get `ON=sqrt(OS^2−SN^2)=sqrt(10^2−6^2)=8 cm`
`OM=sqrt(OQ^2−QM^2)=10^2−8^2cm =6 cm.`
Now two cases arise- (i) `PQ \and\ RS` are to the opposite side of the centre O.
Here,`MN=OM+ON=(6+8)cm=14 cm` (from i)
(ii)` PQ \and\ RS` are to the same side of the center O. (fig II)
Here `MN=ON−OM=(8−6) cm=2 cm.`
So the distance between `PQ \and\ RS =14 cm`
when `PQ \and\ RS` are to the opposite side of the centre `O`
the distance between `PQ \and\ RS =2 cm`
when `PQ \and\ RS` are to the same side of the centre `O.`