The triangular side walls of a flyover have been used for advertisements.

Question:

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?

Solution:

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.

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