The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m, 15 m. The advertisements yield an earning of Rs 2000 per m2 a year. A company hired one of its walls for 6 months. How much rent did it pay?
The sides of the triangle are of length 13 m, 14 m and 15 m.
∴ Semi-perimeter of the triangle is
$s=\frac{13+14+15}{2}=\frac{42}{2}=21 \mathrm{~m}$
∴ By Heron's formula,
Area of $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{21(8)(7)(6)}$
$=84 \mathrm{~m}^{2}$
Now,
The rent of advertisements per m2 per year = Rs 2000
The rent of the wall with area 84 m2 per year = Rs 2000 × 84
= Rs 168000
The rent of the wall with area $84 \mathrm{~m}^{2}$ for 6 months $=$ Rs $\frac{168000}{2}$
= Rs 84000
Hence, the rent paid by the company is Rs 84000.