The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig.). The advertisements yield an earning of ₹ 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

detailed solution

1

We know that, 

The Heron's formula for calculating the area of the triangle,

Area of triangle = $\sqrt{s (s - a) (s - b) (s -c)}$

where, s is the semi-perimeter = half of the perimeter

and  a, b and c are the sides of the triangle 

Triangular sides of walls are, a = 122 m, b = 22 m, c = 120 m

Thus, Semi-perimeter 

s = (a $+$ b $+$ c)/2

= (122 $+$ 22 $+$ 120)/2

= 264/2

= 132 m

Area of a triangular wall = $\sqrt{132 (132 -122) (132-22) (132 -120)}$

= $\sqrt{132 \times 10 \times 110 \times 12}$

= 1320 m

Given that, 

Rent of 1 m² area per year = ₹ 5000

Rent of 1 m² area per month = ₹ 5000/12

Rent of 1320 m² area for 3 months =  (5000/12) $\times$ 3 $\times$1320

= ₹ 1650000

Hence, the company paid ₹ 16,50,000 as rent.