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(A) The sides of the triangular wall are $a = 122\, m, b = 120\, m, c = 22\, m$.The semi-perimeter $s$ is given by:$s = \frac{a + b + c}{2} = \frac{12

Vedclass · Accepted Answer

$16,50,000$

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(A) The sides of the triangular wall are $a = 122\, m, b = 120\, m, c = 22\, m$.The semi-perimeter $s$ is given by:$s = \frac{a + b + c}{2} = \frac{122 + 120 + 22}{2}\, m = \frac{264}{2}\, m = 132\, m$.The area of the triangle using Heron's formula is $\sqrt{s(s-a)(s-b)(s-c)}$.Area $= \sqrt{132(132 - 122)(132 - 120)(132 - 22)}\, m^2$$= \sqrt{132 	imes 10 	imes 12 	imes 110}\, m^2$$= \sqrt{12 	imes 11 	imes 10 	imes 12 	imes 11 	imes 10}\, m^2$$= \sqrt{12^2 	imes 11^2 	imes 10^2}\, m^2 = 12 	imes 11 	imes 10\, m^2 = 1320\, m^2$.Rent for $1$ year ($12$ months) per $m^2 = Rs. 5000$.Rent for $3$ months per $m^2 = Rs. 5000 	imes \frac{3}{12} = Rs. 1250$.Total rent for $1320\, m^2$ for $3$ months $= 1320 	imes 1250 = Rs. 16,50,000$.

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.). 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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