Question

Two chords AB and CD of lengths 5cm and 11cm respectively of a circle are parallel to each other and are on the opposite side of it's center. If the distance between AB and CD is 6cm , find the radius of circle . ​

Answer 1

Join OA and OC.

Let the radius of the circle be r cm and O be the centre 

Draw OP⊥AB and OQ⊥CD. 

We know, OQ⊥CD, OP⊥AB and AB∥CD. 

Therefore, points P,O and Q are collinear. So, PQ=6 cm. 

Let OP=x. 

Then, OQ=(6–x) cm. 

And OA=OC=r. 

Also, AP=PB=2.5 cm and CQ=QD=5.5 cm.

(Perpendicular from the centre to a chord of the circle bisects the chord.)

  

In right triangles QAP and OCQ, we have

OA2=OP2+AP2 and OC2=OQ2+CQ2

∴r2=x2+(2.5)2                 ..... (1) 

and r2=(6−x)2+(5.5)2      ..... (2) 

⇒x2+(2.5)2=(6−x)2+(5.5)2

⇒x2+6.25=36−12x+x2+30.25

12x=60

∴x=5

Putting x=5 in (1)

r² = 5² + (2.5)² = 25 + 6.25 = 31.25

r² = 31.25 = 5.6 cm

the radius of the circle is 5.6 cm