Question

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

Answer 1

In right angled ∆ABD,
AB² = AD² + DB² (Pythagoras Theorem)
⇒ AB² = 12² + 16²
⇒ AB² = 144 + 256
⇒ AB² = 400
⇒ AB = 20 cm

Area of ∆ADB = $\frac{1}{2}\times DB\times AD$
= $\frac{1}{2}\times 16\times 12$
= 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,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ACB=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{60\left(60-20\right)\left(60-52\right)\left(60-48\right)} \\ =\sqrt{60\left(40\right)\left(8\right)\left(12\right)} \\ =480{\mathrm{cm}}^2...\left(2\right) \end{gathered}$$


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.