Two circle of radii 14 cm and 10 cm intersect each other, and the length of their common chord is 16 cm. What is the distance (in m) between their centres?

Let P and Q are the centre of the circle of Radii 14 and 10 respectively.

AB = 16 is the common chord, (Given)

We have AP = 14 cm

AQ = 10 cm

AM = $$\frac{AB}{2}=\frac{16}{2}=8\ cm$$

In a Right angled $$\triangle\ $$ APM

$$AP^2=AM^2+PM^2$$

$$PM^2=196-64=132$$

PM = $$\sqrt{\ 132}$$ cm = $$2\sqrt{\ 33}$$ cm

$$\longrightarrow\ $$ In $$\triangle\ $$ AQM

$$AQ^2=AM^2+MQ^2$$

$$MQ^2=100-64=36$$

$$MQ=6$$ cm

$$\therefore\ PQ\ =\ PM+MQ$$ 

$$\therefore\ 2\sqrt{\ 33}+6$$ = $$\ 6+2\sqrt{\ 33}$$

Hence, Option B is correct.