**AB, CD are parallel chords of a circle 7 cm apart. If AB = 6 cm, CD = 8 cm, find the radius of the circle.**

###### Answers:

Let O be the centre of the circle OM and ON are perpendicular on AB and CD.

MON is one straight line.

Here AM = ½ AB = 3 cm, CN = ½ CD = 4 cm

Let ON = x cm and radius OA = OC = r cm

From right angled triangle OCN,

ON^{2} = OC^{2} – CN^{2} [By Pythagoras Theorem]

x^{2} = r^{2} – 16 ….(1)

From right angled triangle OAM,

OM^{2} = OA^{2} – AM^{2 } [By Pythagoras Theorem]

(7 – x)^{2} = r^{2} – 9 …..(2)

From (1) and (2)

(7 – x)^{2} – x^{2} = 7

49 + x^{2} – 14x – x^{2} = 7

14x = 42

x = 3

from (1), r^{2} = x^{2} + 16

= 9 + 16 = 25

r = 5 cm

Hence, the radius of the circle is 5 cm.