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# 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?

"

Given:

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 $\ Rs. 1650000$.

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