NCERT Solutions
Class 9 Maths
Circles

Ex.10.6 Q.2
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other
and are on opposite sides of its centre.
If the distance between AB and CD is 6 cm, find the radius of the circle.
Let O be the centre of the circle and let its radius be r cm.
Draw OM ⊥ AB and OL ⊥ CD.
Then, AM = =
cm
and, CL = =
cm
Since, AB || CD, it follows that the points O, L, M are collinear and therefore, LM = 6 cm.
Let OL = x cm. Then OM = (6 – x) cm
Join OA and OC. Then OA = OC = r cm.
Now, from right-angled ∆OMA and ∆OLC, we have
OA2 = OM2 + AM2 and OC2 = OL2 + CL2
[By Pythagoras Theorem]
=> r2 = (6 – x) 2 + ( ) 2 ……... (1)
and r2 = x2 + ( ) 2 ………... (2)
From equation (1) and (2), we get
=> (6 – x) 2 + ( ) 2 = x2 + (
) 2
=> 36 + x2 – 12x + = x2 +
=> -12x = –
– 36
=> -12x = – 36
=> -12x = 24 – 36
=> 12x = – 12
=> x = 1
Substituting x = 1 in equation (1), we get
r2 = (6 – x) 2 + ( ) 2
=> r2 = (6 – 1) 2 + ( ) 2
=> r2 = 52 +
=> r2 = 25 +
=> r2 =
=> r = √ ( )
=> r =
Hence, radius r = cm.