Question

AB and CD are two chords of a circle such that AB = 6 cm, CD = 12 cm and AB || CD. If the distance between AB and CD is 3 cm, find the radius of the circle.

Answer 1

Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm.
Let the radius of the circle be r cm,
Draw OP ⊥ AB and OQ ⊥ CD.
Since AB || CD and OP ⊥ AB, OQ ⊥ CD. 
Therefore, points O, Q and P are collinear. 
Clearly, PQ = 3 cm.

Let OQ = x cm. Then, OP = (x + 3) cm
In right triangles OAP and OCQ, we have
OA² = OP² + AP² and OC² = OQ² + CQ²
⇒ r² = (x + 3)² + 3² and r² = x² + 6²

   ....[ ∵ AP = `1/2"AB" = 3 "cm" and CQ = 1/2"CD = 6 cm`]

⇒ (x + 3)² + 3² = x² + 6²  ...(On equating the value of r²)
⇒ 6x = 18
⇒ x = 3 cm

Putting the values of x in r² = x² + 6², we get
r² = x² + 6² = 45
⇒ r = `sqrt45` cm = 6.7 cm
Hence the radius of the circle is 6.7 cm