In ΔABC, ∠B = 90° (Given)
AB = 9 cm, AC = 15 cm (Given)
By using Pythagoras theorem
AC2 = AB2 + BC2
⇒152 = 92 + BC2
⇒BC2 = 225 – 81
= 144
or BC = 12
Again,
AD = DB = AB/2 = 9/2 = 4.5 cm [D is the mid−point of AB
D and E are mid-points of AB and AC
DE ∥ BC ⇒ DE = BC/2 [By mid−point theorem]
Now,
Area of ΔADE = 1/2 x AD x DE
= 1/2 x 4.5 x 6
= 13.5
Area of ΔADE is 13.5 cm2