Question

Two chords AB and CD of lengths 6 cm and 14 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 8 cm, find the radius of the circle.​

Answer 1

Answer:

GIVE BRAINLIEST

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

OA  

2

=OP  

2

+AP  

2

 and OC  

2

=OQ  

2

+CQ  

2

 

∴r  

2

=x  

2

+(2.5)  

2

                 ..... (1)  

and r  

2

=(6−x)  

2

+(5.5)  

2

      ..... (2)  

⇒x  

2

+(2.5)  

2

=(6−x)  

2

+(5.5)  

2

 

⇒x  

2

+6.25=36−12x+x  

2

+30.25

12x=60

∴x=5

Putting x=5 in (1), we get  

r  

2

=52+(2.5)  

2

=25+6.25=31.25

⇒r  

2

=31.25⇒r=5.6

Hence, the radius of the circle is 5.6 cm

Step-by-step explanation: