The difference between the two adjoining sides of a triangle is $ 20 \mathrm{~cm} $, third side is $ 40 \mathrm{~cm} $ and the perimeter of the same triangle $ 120 \mathrm{~cm} $. Find the area of the triangle.

Given:

The difference between the two adjoining sides of a triangle is \( 20 \mathrm{~cm} \), third side is \( 40 \mathrm{~cm} \) and the perimeter of the same triangle \( 120 \mathrm{~cm} \).

To do:

We have to find the area of the triangle.

Solution:
 Let one of the two sides be $x$

This implies,

The second side $=20+x$

Perimeter of the triangle $=x+20+x+40=120$

$2x+60=120$

$2x=120-60$

$x=\frac{60}{2}$

$x=30$

$\Rightarrow 20+x=20+30=50$

Here,

$30^2+40^2=900+1600=2500$

$50^2=2500$

This implies,

The given triangle is aright angled triangle.

Therefore,

The area of the given triangle $=\frac{1}{2}\times30\times40$

$=15\times40$

$=600\ cm^2$

The area of the given triangle is $600\ cm^2$.