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The triangular side walls of a flyover have been used for advertisements. 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 122 m, 22 m, and 120 m respectively.

Perimeter of triangle = (122 + 22 + 120) m

2*s* = 264 m

*s* = 132 m

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 = 1320m^2`

Rent of 1 m^{2} area per year = Rs. 5000

Rent of 1 m^{2} area per month = Rs. 5000/12

Rent of 1320 m^{2} area for 3 months

`= Rs (5000/12xx3xx1320)`

= Rs (5000 × 330)

= Rs 1650000

Therefore, the company had to pay Rs 1650000.

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