Question

Two chords of lengths $30$cm and $16$cm are on the opposite side of the centre of the circle. If the radius of the circle is $17$cm, find the distance between the chords.

Answer 1

Let $AB$ $=30$$cm$ and $CD$ $=16$ $cm$, $r=OC=OA=7$ $cm.$
Join $OA$ and $OC$.

Since the altitudes from the center to the chord bisect the chord, therefore M and N are the mid-points of AB and CD respectively.

$AM=\frac{1}{2}AB=\frac{1}{2}\times 30=15$cm.

$CN=\frac{1}{2}CD=\frac{1}{2}\times 16=8$cm

In right angle $\Delta OMA$,
$O{M}^2=O{A}^2-A{M}^2$
$=(17{)}^2-(15{)}^2$
$=289-225=64=(8{)}^2$
$OM=8cm$

In right angle $\Delta ONC$
$O{N}^2=O{C}^2-N{C}^2$
$=(17{)}^2-(8{)}^2$
$=289-64$
$=225$
$=(15{)}^2$
$\therefore ON=15cm$
Now, $MN=OM+ON$
$=8+15=23$cm
Thus, the distance between the chords is $23$cm.
$\therefore MN=23cm$