Question

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 as 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 ?

• A. .Rs.$16,500,000$
• B. .Rs.$17,500,000$
• C. .Rs.$16,000,000$
• D. .none of the above

Answer 1

The correct option is A .Rs.$16,500,000$

Refer figure,

In $△ABC$

$a=122m,b=22m,c=120m$

$s=\frac{a+b+c}{2}=\frac{122+22+120}{2}$

$=132m$

$ar△ABC=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{132(132-122)(132-23)(132-120)}$

$=\sqrt{132\times 10\times 110\times 12}=\sqrt{132\times 10\times 11\times 10\times 12}=\sqrt{{\left(132\right)}^2\times {\left(10\right)}^2}$

$=132\times 10=1320m^2$

Rent of $1m^2$ per year = Rs. $5000$

Rent of $1m^2$ per month = Rs.$\frac{5000}{12}$

Rent of $1m^2$ for $3$ months=Rs.$\frac{3\times 5000}{12}$

Rent of $1320m^2$ for $3$ months=Rs. $\left(\frac{3\times 5000\times 1320}{12}\right)$

= Rs.$16,500,000$