Question

The triangular side walls of a flyover have been used for advertisement. The sides of the walls are 122 m, 22 m, and 120 m (see the given figure). The advertisements yield an earning of Rs 5000 per $m^2$ per year. A company hired one of its walls for 3 months. How much rent did it pay?

Answer 1

Let the sides of the triangle (i.e., a, b, c) be 122 m, 22 m, and 120 m respectively.
Perimeter of triangle = (122 + 22 + 120) m
2s = 264 m
∴ s = 132 m

By Heron’s formula,
Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area of the given triangle}=\sqrt{132(132-122)(132-22)(132-120)}{m}^2$
$=\sqrt{132\left(10\right)\left(110\right)\left(12\right)}{m}^2=1320m^2$
Rent of 1 $m^2$ area per year = Rs 5000
Rent of 1 $m^2$ area per month = $Rs\frac{5000}{12}$
Rent of 1320 $m^2$ area for 3 months = $Rs(\frac{5000}{12}\times 3\times 1320)=Rs(5000\times 330)=Rs16,50,000$

Therefore, the company paid Rs 16,50,000.