The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $ 122 \mathrm{~m}, 22 \mathrm{~m} $ and $ 120 \mathrm{~m} $ (see Fig. 12.9). The advertisements yield an earning of ₹ 5000 per $ \mathrm{m}^{2} $ per year. A company hired one of its walls for 3 months. How mech rent did it pay? "
The sides of the triangular walls of the flyover are $122\ m, 22\ m$ and $120\ m$. The advertisements yield an earning of $\ RS. 5000\ per\ m^2$ per year. A company hired one of its walls for 3 months.
To find:
We have to find the rent paid by the company.
Solution:
The sides of the triangular walls of the flyover are $122\ m, 22\ m$ and $120\ m$.
We know that,
Perimeter $P$ of a triangle with sides of length $a$ units, $b$ units and $c$ units
$P=(a+b+c)$units.
This implies,
$P=122\ m+22\ m+120\ m$
$P=264\ m$
Now, we have perimeter as $264\ m$.
By Heron's formula:
$A=\sqrt{s(s-a)(s-b)(s-c)}$
Since,
$S=\frac{a+b+c}{2}$
$S=\frac{122+22+120}{2}$
$S=\frac{264}{2}$
$S=132\ m$
This implies,
$A=\sqrt{132(132-122)(132-22)(132-120)}$
$A=\sqrt{132(10)(110)(12)}$
$A=1320\ m^2$
Therefore,
The area of the triangular wall of the flyover is $1320\ m^2$.
We have,
The advertisements yield of the triangular side of the walls per year $=\ RS. 5000\ per\ m^2$
This implies,
The advertisements yield of the triangular side of the walls per one month$=\frac{\ Rs.5000\ per\ m^2}{12}$
$=416.66\ m^2$
Therefore,
The rent paid by the company for one of its walls of area $1320\ m^2$ for 3 months $=416.66\ m^2\times3\times1320\ m^2$
$=\ Rs. 1650000$.
Hence,
The rent paid by the company for one of its walls of area $1320\ m^2$ for 3 months is $\ Rs. 1650000$.