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The triangular side walls of a flyover have been used for advertisements. The sides of the wails are 122 m, 22 m and 120 m (see figure). The advertisements yield earnings 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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Answer :

Let a = 122 m, b = 22 m, c = 120 m
Also, we have,

\(b^2 + c^2 = (22)^2 + (120)^2 = 484 + 14400 = 14884 = (122)^2 = a^2\)

Thus, we observe that the side walls are in right triangular shape.

Thus, the area of the triangular side walls
= \(\frac{1}{2} × a × c\)
\(\Rightarrow \)= \(\frac{1}{2} × 22 × 120\)
\(\Rightarrow \) = \(11 × 120 = 1320 m^2\)

We know that, yearly rent = Rs.5000 per \(m^2\)

Therefore, yearly rent
= Rs.5000 × \(\frac{1}{2} \) per \( m^2\)

Now, the company has hired one of its walls for 3 months.

Thus, rent paid by the company for 3 months
= 1320 × \(\frac{5000}{12}\) × 3
= 110 × 5000 × 3
= Rs. 1650000

Therefore, rent paid by the company for 3 months = Rs. 1650000.