The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig.).
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig.). The advertisements yield an earning of ₹5000 per m^{2} per year. A company hired one of its walls for 3 months. How much rent did it pay?
The sides of the triangle ABC are 122 m, 22 m and 120 m respectively.
Now, the perimeter will be (122+22+120) = 264 m
Also, the semi perimeter (s) = 264/2 = 132 m
Using Heron’s formula,
Area of the triangle = \(\sqrt{s(s  a)(s  b)(s  c)}\)
= \(\sqrt{132(132  122)(132  22)(132  120)m^2}\)
= \(\sqrt{132 \times 10 \times 110 \times 12 m^2}\)
=1320 m^{2}
We know that the rent of advertising per year = ₹ 5000 per m^{2}
∴ The rent of one wall for 3 months
= Rs. (1320 × 5000 × 3)/12
= Rs 1650000

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