Question:
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre?
Solution:
Distance of smaller chord AB from centre of circle = 4 cm, OM = 4 cm
MB = AB/2 = 6/2 = 3 cm
In ΔOMB
$\mathrm{OM}^{2}+\mathrm{MB}^{2}=\mathrm{OB}^{2}$
$4^{2}+9^{2}=\mathrm{OB}^{2}$
$16+9=\mathrm{OB}^{2}$
$O B=\sqrt{25}$
OB = 5 cm
In ΔOND
OD = OB = 5 cm [Radii of same circle]
ND = CD/2 = 8/2 = 4 cm
$O N^{2}+N D^{2}=O D^{2}$
$O N^{2}+4^{2}=5^{2}$
$O N^{2}=25-16$
$O N=\sqrt{9}$
ON = 3 cm
So, the distance of bigger chord from the circle is 3 cm.