Question

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $$ 122 m, 22m $$ and $$ 120m$$. 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?

Answer 1

Sides of triangle $$ a = 122 m , b= 22 m, c= 120 cm $$
$$ \therefore s = \dfrac {a+b+c}{2} $$
$$ = \dfrac {122 +22 +120}{2} $$
$$ =\dfrac {264}{2} $$
$$ s = 132 m $$
$$ \therefore $$ area of triangle A
$$ A = \sqrt { s(s-a)(s-b)(s-c) } $$
$$ = \sqrt{ 132 (132 -122)(132-22)(132-120) } $$
$$ = \sqrt { 132 \times 10 \times 110 \times 12 } $$
$$ = \sqrt {1320 \times 1320} $$
$$ = \sqrt { (1320)^2 } $$
$$ \therefore A = 1320 sq.cm $$
Rent for each year, each sq m =$$ rs 5000 $$
Rent for each month each sq.m = $$ \frac {5000}{12} $$
Rent for each quarterly, each sq.m $$ = 3 \times \frac {5000}{12} = rs 1250 $$
$$ \therefore $$ Rent for $$ 1320 sq. m =1250 \times 1320 =Rs . 1650000 $$
$$ \therefore $$ Amount company has to pay is $$ rs . 1650000 $$.