In a circle of radius 10 cm, PQ and RS are two parallel chords of lengths 16 cm and 12 cm, respectively. What is the distance between the chords if they are on opposite sides of the centre?

Given,

Radius of the circle = 10 cm

From the figure,

In $$\triangle\ $$OBS,

$$OB^2+BS^2=OS^2$$

$$=$$>  $$OB^2+6^2=10^2$$

$$=$$>  $$OB^2+36=100$$

$$=$$>  $$OB^2=64$$

$$=$$>  $$OB=8$$ cm

In $$\triangle\ $$OPA,

$$OA^2+AP^2=OP^2$$

$$=$$> $$OA^2+8^2=10^2$$

$$=$$> $$OA^2+64=100$$

$$=$$> $$OA^2=36$$

$$=$$> $$OA=6$$ cm

$$\therefore\ $$Distance between the parallel chords = AB = OA + OB = 6 + 8 = 14 cm

Hence, the correct answer is Option A