Question 2
The triangular side walls of a flyover have been used for advertisement. The sides of the walls are 122m, 22m, and 120m (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?

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Solution

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

By Heron’s formula,
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
$Area of given triangle = \sqrt{132 (132 - 122) (132 - 22) (132 - 120)} m^{2}$
$= \sqrt{132 (10) (110) (12)} m^{2} = 1320 m^{2}$
Rent of 1 $m^{2}$ area per year = Rs 5000
Rent of 1 $m^{2}$ area per month = $R s \frac{5000}{12}$
Rent of 1320 $m^{2}$ area for 3 months = $R s (\frac{5000}{12} \times 3 \times 1320) = R s (5000 \times 330) = R s 1650000$

Therefore, the company has to pay Rs 1650000.

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The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122 m, 22 m, 120 m$ . The advertisements yield an earning of $R s . 5000$ per $m^{2}$ per year.. A company hired one of its walls for $3$ months. How much rent did it pay?