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 figure below). The advertisements yield an earning of ₹ 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
Solution and Explanation
The sides of the triangle (i.e., a, b, c) are of 122 m, 22 m, and 120 m respectively.
Perimeter of triangle = (122 + 22 + 120) m
2s = 264 m
s = 132 m
By Heron's formula
Area of a triangle =\( \sqrt{\text{s(s - a)(s - b)(s - c)}}\)
= \(\sqrt{\text{132(132 - 122) (132 - 22) (132 - 120)}}\)
\(= \sqrt{\text{132 × 10 × 110 × 12}}\)
= 1320 m2
Rent of 1 m2 area per year = ₹ 5000
Rent of 1 m2 area per month = \(₹ \frac{5000}{12}\)
Rent of 1320 m2 area for 3 months = \(₹ \)(\(\frac{5000}{12}\)) × 3 × 1320
= \(₹\)1650000
Therefore, the company paid ₹ 16,50,000 as rent.
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