Let O be the centre of the circle and let its radius be r cm.

Draw OM ⊥ AB and OL ⊥ CD.

Then, AM = 1/ 2 AB = 5/ 2 cm

and, CL = 1/ 2 CD = 11/ 2 cm

Since, AB || CD, it follows that the points O, L, M are 10 CIRCLES

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

OA^{2} = OM^{2 }+ AM^{2} and OC^{2} = OL^{2} + CL^{2} [By Pythagoras Theorem]

⇒ r^{2} = (6 – x)^{2} + (5/ 2) ^{2} ..(i) and r^{2} = x^{2} + (11/ 2)^{ 2} ... (ii)

⇒ (6 – x)^{2} + (5/ 2)^{ 2} = x^{2} + (11/ 2)^{ 2} [From (i) and (ii)]

⇒ 36 + x2 – 12x + 25/ 4 = x2 + 121/ 4

⇒ – 12x = 121/ 4 – 25/ 4 – 36

⇒ – 12x = 96/ 4 – 36

⇒ – 12x = 24 – 36

⇒ – 12x = – 12

⇒ x = 1

Substituting x =1 in (i), we get

r^{2} = (6 – x)2 + (5/ 2) ^{2 }

⇒ r^{2} = (6 – 1)2 + (5/ 2)^{ 2 }

⇒ r^{2} = (5)2 + (5/ 2) 2= 25 + 25/ 4

⇒ r^{2} = 125/ 4

⇒ r = (5 \5) 2

Hence, radius r = (5\ 5) 2 cm. Ans.