Question

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122m,22m,120m$. 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

In triangular side walls of a flyover,

The sides of the walls are , $\left(a\right)=122m,\left(b\right)=22m,\left(c\right)=120m$

$S=\frac{a+b+c}{2}$ [Semi perimeter]

$\Rightarrow S=\frac{120m+122m+22m}{2}$

$=\frac{264m}{2}$

$=\frac{264m}{2}$

$\therefore S=132m$

According to Heron's formula,

Area of triangle with the side $a,b$ and $c$ is

$△=\sqrt{S(S-a)(S-b)(S-c)}$, where, $S=\frac{a+b+c}{2}$

$=\sqrt{132m(132m-122m)(132m-120m)(132m-22m)}$

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

$=\sqrt{11\times 12\times 10\times 12\times 11\times 10}{m}^2$

$=(11\times 10\times 12)m^2$

$=1320m^2$

Therefore, Area of the triangular side walls of a flyover is $1320m^2$.

It is given that,

Rent of $1m^2$ per year $=Rs.5000$

$\Rightarrow$ Rent of $1m^2$ per month $=\frac{Rs.5000}{12}$

$\Rightarrow$ Rent of $1m^2$ per $3$ months $=3times\frac{Rs.5000}{12}=\frac{Rs.5000}{4}$

$\Rightarrow$ Rent of $1320m^2$ per $3$ months

$=1320\times \frac{Rs.5000}{4}$

$=Rs.5000\times 330$

$=Rs.1650000$

Therefore, The company had to pay is $Rs.16,50,000$