The triangular side walls of a flyover have been

Since, the sides of a triangular walls are $a=13 \mathrm{~m}, b=14 \mathrm{~m}$ and $c=15 \mathrm{~m}$

$\therefore$ semi-perimeter of triangular side wall, $s=\frac{a+b+c}{2}=\frac{13+14+15}{2}=\frac{42}{2}=21 \mathrm{~m}$

$\therefore$ Area of triangular side wall $=\sqrt{s(s-a)(s-b)(s-c)}$   [by Heron's formula]

$=\sqrt{21(21-13)(21-14)(21-15)}$

$=\sqrt{21 \times 8 \times 7 \times 6}$

$=\sqrt{21 \times 4 \times 2 \times 7 \times 3 \times 2}$

$=\sqrt{(21)^{2} \times(4)^{2}}=21 \times 4=84 \mathrm{~m}^{2}$

Since, the advertisement yield earning per year for $1 \mathrm{~m}^{2}=₹ 2000$

$\therefore$ Advertisement yield earning per year on $84 \mathrm{~m}^{2}=2000 \times 84=₹ 168000$

As the company hired one of its walls for 6 months, therefore company pay the rent

$=\frac{1}{2}(168000)=₹ 84000$

Hence, the company paid rent ₹ 84000 .