Chords AB and CD of a circle intersect externally at P. If AB = 6 cm, CD = 3 cm and PD = 5 cm, then the length of PB is

AB = 6 cm; CD = 3 cm

PD = 5 cm; PB = ?

PA × PB = PC × PD

$\Rightarrow(\mathrm{PB}-6) \mathrm{PB}=2 \times 5$

$\Rightarrow \mathrm{PB}^{2}-6 \mathrm{~PB}-10=0$

$\Rightarrow \mathrm{PB}=\frac{6 \pm \sqrt{36+40}}{2}=\frac{6 \pm \sqrt{76}}{2}=\frac{6+8.7}{2} \approx 7.35$