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Q2.    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. 12.9). The advertisements yield an earning of  Rs. 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?


Answers (1)


From the figure, 

The sides of the triangle are:

a = 122m,\ b = 120m\ and\ c = 22m

The semi perimeter, s will be

s = \frac{a+b+c}{2} = \frac{122+120+22}{2} = \frac{264}{2} = 132m

Therefore, the area of the triangular side wall will be calculated by the Heron's Formula,

A = \sqrt{s(s-a)(s-b)(s-c)}

     = \sqrt{132(132-122)(132-120)(132-22)}\ m^2

     = \sqrt{132(10)(12)(110)}\ m^2

     = \sqrt{(12\times11)(10)(12)(11\times10)}\ m^2 = 1320\ m^2

Given the rent for 1 year (i.e., 12 months) per meter square is Rs. 5000.

Rent for 3 months per meter square will be: 

Rs.\ 5000\times \frac{3}{12}

Therefore,  for 3 months for 1320 m:

Rs.\ 5000\times \frac{3}{12}\times 1320 = Rs.\ 16,50,000.



Posted by

Divya Prakash Singh

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