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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.

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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 

OA2 = OM2 + AM2 and OC2 = OL2 + CL2 [By Pythagoras Theorem] 

⇒ r2 = (6 – x)2 + (5/ 2) 2 ..(i) and r2 = x2 + (11/ 2) 2 ... (ii)

⇒ (6 – x)2 + (5/ 2) 2 = x2 + (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 

r2 = (6 – x)2 + (5/ 2) 2  

⇒ r2 = (6 – 1)2 + (5/ 2) 

⇒ r2 = (5)2 + (5/ 2) 2= 25 + 25/ 4 

⇒ r2 = 125/ 4 

⇒ r = (5 \5) 2 

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

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