The triangular side walls of a flyover have been used for advertisements. The sides of walls are$122text{ m, 22m }!!And!!text{ 120m}$ (see figure 12.9). The advertisements yield the earning of Rs 5000 per${{text{m}}^{2}}$ year. A company hired one of its walls for 3 months. How much rent did it pay?

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Rs.16,500,000

Rs.17,000,000

16,000,000

None of these

answer is A.

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Detailed Solution

Refer figure,
In△ABC
a = 122m, b=22m, c = 120m
s = (a+b+c)/2 = (122+22+120)/2
=132m
ar△ABC = √ s(s-a)(s-b)(s-c) = √ 132(132 - 122)(132 - 23)(132 - 120)
= √ 132 ✕ 10 ✕ 110 ✕ 12 = √ 132 ✕ 10 ✕ 11 ✕10 ✕ 12 = √ (132)^2 ✕ (10)^2
= 132 ✕ 10 = 1320 m^2
Rent of 1 m^2 per year = Rs. 5000
Rent of 1 m^2 per month = Rs. 5000/12
Rent of 1 m^2 for 3 per month = Rs. (3 ✕ 5000)/12
Rent of 1320 m^2 for 3 months = Rs. (3 ✕ 5000 ✕ 1320)/12
= Rs. 16,500,000

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