Remember: The perpendicular drawn from the centre of the circle to the chord bisects the chord. Now, OE ⊥ AB ⇒AE=BE =
AB
2
=2cm Also, OF ⊥ CD ⇒ CF = DF =
CD
2
=5cm EF = 3cm In right ΔOCF, OF2=OC2−CF2=r2−25....(1) In right ΔOAE, OE2=OA2−AE2=r2−4....(2) Subtracting (1) from (2), we have (OF+3)2−OF2=21 ⇒6OF=21−9=12 ⇒OF=2cm Putting OF = 2 in (1), we have r2=4+25=29 ⇒r=√29cm