Question

The shortest distance between two parallel chords of length 32 units and 24 units of circle having radius 20 units is units

Answer 1

The position of the two parallel chords having minimum distance between them is shown in the following figure:

P is the centre of the circle.
PR bisects both the chords at Q and R, respectively, and it is perpendicular to both the chords.

Hence, $XQ=\frac{32}{2}=16\text{units}$
XP is the radius of the triangle.

For right triangle $△XPQ,$
$PQ=\sqrt{X{P}^2-X{Q}^2}$
$\Rightarrow PQ=\sqrt{{20}^2-{16}^2}=12\text{units}$

Similarly for $△UPR,$
$PR=\sqrt{U{P}^2-U{R}^2}\Rightarrow PR=\sqrt{{20}^2-{12}^2}=16\text{units}$

$QR=PR-PQ=16-12=4\text{units}$

$\therefore$ The required distance is $4\text{units}.$