Question

Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle.

Answer 1

Let r be the radius of the given circle and its center be O. 

Draw OM ⊥ AB and ON⊥ CD. 

Since, 

OM perpendicular AB, 

ON perpendicular CD.

And, 

AB || CD

Therefore, 

Points M, O and N are collinear. 

So, 

MN = 6cm 

Let, 

OM = x cm. 

Then,

ON = (6 - x)cm. 

Join OA and OC. 

Then, 

OA = OC = r 

As the perpendicular from the centre to a chord of the circle bisects the chord. 

∴ AM = BM = 1/2 AB 

= 1/2 x 5 = 2.5cm 

CN = DN = 1/2CD 

= 1/2 x 11 = 5.5cm 

In right triangles OAM and OCN, 

We have, 

OA² = OM²+ AM² 

And,

OC²= ON² + CN²

From (i) and (ii), we have

⇒ 4x² + 25 = 144 + 4x²- 48x + 121 

⇒ 48x = 240 

⇒ x = 240/48

⇒ x = 5 

Putting the value of x in euation (i), we get 

r² = 52 + (5/2)² 

⇒ r² = 25 + 25/4 

⇒ r²= 125/4 

⇒ r = 5√5/2 cm