Find the area of the shaded region in the figure given below.

Solution:

In right angled ∆ABD,

$A B^{2}=A D^{2}+D B^{2}$    (Pythagoras Theorem)

$\Rightarrow A B^{2}=12^{2}+16^{2}$

$\Rightarrow A B^{2}=144+256$

$\Rightarrow A B^{2}=400$

$\Rightarrow A B=20 \mathrm{~cm}$

Area of $\triangle A D B=\frac{1}{2} \times D B \times A D$

$=\frac{1}{2} \times 16 \times 12$

= 96 cm2               ....(1)

In ∆ACB,
The sides of the triangle are of length 20 cm, 52 cm and 48 cm.
∴ Semi-perimeter of the triangle is

$s=\frac{20+52+48}{2}=\frac{120}{2}=60 \mathrm{~cm}$

∴ By Heron's formula,

Area of $\Delta A C B=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{60(60-20)(60-52)(60-48)}$

$=\sqrt{60(40)(8)(12)}$

$=480 \mathrm{~cm}^{2} \quad \ldots(2)$

Now,
Area of the shaded region = Area of ∆ACB − Area of ∆ADB
= 480 − 96
= 384 cm2

Hence, the area of the shaded region in the given figure is 384 cm2.