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 advertisement. The sides of the walls are 122m, 22m, and 120m (see the given figure). The advertisements yield an earning of Rs 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

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?