Question

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

• A. 96 cm${}^2$
• B. 576 cm${}^2$
• C. 480 cm${}^2$
• D. 384 cm${}^2$

Answer 1

The correct option is D 384 cm${}^2$

$\because \Delta$ ABD, is a right angled triangle
$\therefore A{B}^2=A{D}^2+D{B}^2$
$\Rightarrow A{B}^2={12}^2+{16}^2$
$\Rightarrow A{B}^2=144+256$
$\Rightarrow A{B}^2=400$
$\Rightarrow$ AB = 20 cm
Area of $\Delta$ ADB = $\frac{1}{2}$ × DB × AD
= 0.5 × 16 × 12
= 96 cm${}^2$
In ∆ ACB,
The sides of the triangle are of length 20 cm, 52 cm and 48 cm.
∴ Semi-perimeter of the triangle is:
$\text{Semi-Perimeter (s)}=\frac{20+52+48}{2}=\frac{120}{2}=60\text{cm}$
$\therefore$ By Heron's formula,
Area of $\Delta$ ACB is:
=$\sqrt{s(s-a)(s-b)(s-c)}$
=$\sqrt{60(60-20)(60-52)(60-48)}$
= $\sqrt{60\left(40\right)\left(8\right)\left(12\right)}$
= 480 cm${}^2$
Now,
Area of the shaded region = Area of $\Delta$ ACB − Area of $\Delta$ ADB
= 480 − 96
= 384 cm${}^2$
Hence, the area of the shaded region in the given figure is 384 cm${}^2$.