In ∆ABC, if AB = 16 cm, BC = 12 cm and AC = 20 cm, then ∆ABC is

(b) right-angled

We have:

$A B^{2}+B C^{2}=16^{2}+12^{2}=256+144=400$

and, $A C^{2}=20^{2}=400$

$\therefore A B^{2}+B C^{2}=A C^{2}$

Hence, ∆ABC is a right-angled triangle.