In figure, Triangle ABC is a right angled triangle at B.

In ΔABC, ∠B = 900,

By using Pythagoras theorem

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

$\Rightarrow 15^{2}=9^{2}+B C^{2}$

$\Rightarrow B C=\sqrt{15^{2}-9^{2}}$

$\Rightarrow \mathrm{BC}=\sqrt{225-81}$

$\Rightarrow \mathrm{BC}=\sqrt{144}=12 \mathrm{~cm}$

In ΔABC,

D and E are mid-points of AB and AC

∴ DE ∥ BC, DE = (1/2) BC [By mid−point theorem]

AD = DB = AB/2 = 9/2 = 4.5 cm [∵ D is the mid−point of AB]]

Area of ΔADE = 1/2 ∗AD∗DE

= 1/2 ∗4.5∗6

$=13.5 \mathrm{~cm}^{2}$