In a ΔABC, the bisector of ∠A meets BC at D. If AB = 9.6 cm, AC = 11.2 cm and BD = 4.8 cm, the perimeter (in cm) of ΔABC is :

30.4 28.6 31.2 32.8

31.2

\(\frac{AB}{CD}\) = \(\frac{AB}{AC}\) \(\frac{4.8}{CD}\)= \(\frac{9.6}{11.2}\) = CD = 5.6 So, BC = 4.8 + 5.6 = 10.4 Now, Perimeter of ΔABC = 9.6 + 10.4 + 11.2 = 31.2 cm