AB and CD are two parallel chords of a circle such that AB = 6 cm and CD = 2AB. Both chords are on the same side of the center of the circle. If the distance between them is equal to one-fourth of the length of CD, then the radius of the circle is:

Given chords AB=6 cm, CD =12 cm and AB||CD

Draw OP⊥ AB. Let it intersect CD at Q and AB at P

∴ AP = PB = 3 cm and CQ = DQ = 6 cm [Since perpendicular draw from the center of the chord bisects the chord]

Let OD = OB = r

In right $$\triangle OQD,r^2 =x^2 +6^2$$  [By Pythagoras theorem]

$$\Rightarrow r^2 =x^2 +36---------(i)$$

In right $$\triangle OPB, r^2= (x+3)^2 +3^2$$ [By Pythagoras theorem]

$$\Rightarrow r^2=x^2+ 6x + 9 + 9 =x^2+ 6x + 18------(ii)$$

From (1) and (2) we get

$$\Rightarrow x^2+36=x^2+6x+18$$

$$\Rightarrow 6x=18$$

$$\Rightarrow x=3$$

Put x = 3 in (1), we get

$$\Rightarrow r^2=3^2+36=9+36=45$$

$$\Rightarrow r=\sqrt{45}=3\sqrt{5}$$