Chords AB and CD of a circle intersect externally at P. If AB = 6 cm, CD = 3 cm and PB = 4 cm, then the length (in cm) of PD is:

Given,

AB = 6 cm, CD = 3 cm and PB = 4 cm

Let PD = $$x$$ cm

If chords AB and CD of a circle intersect externally at P then $$\text{PA}\times \text{PB}=\text{PC}\times \text{PD}$$

$$=$$>  $$\left(AB+PB\right)\times PB=\left(CD+PD\right)\times PD$$

$$=$$>  $$\left(6+4\right)\times4=\left(3+x\right)\times x$$

$$=$$>  $$40=3x+x^2$$

$$=$$>  $$x^2+3x-40=0$$

$$=$$>  $$x^2+8x-5x-40=0$$

$$=$$>  $$x\left(x+8\right)-5\left(x+8\right)=0$$

$$=$$>  $$\left(x+8\right)\left(x-5\right)=0$$

$$=$$>  $$x+8=0$$  or   $$x-5=0$$

$$=$$>  $$x=-8$$   or  $$x=5$$

$$x$$ cannot be negative

$$=$$>  $$x=5$$

$$\therefore\ $$PD = $$x$$ = 5 cm

Hence, the correct answer is Option A