Question

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are$13\mathrm{m},14\mathrm{m},15\mathrm{m}.$ The advertisements yield an earning of$₹2000\mathrm{per}{\mathrm{m}}^2$ a year. A company hired one of its walls for$6$ months. How much rent did it pay?

Answer 1

Step1:Find the semi-perimeter of triangle.

Let $\mathrm{a}=13,\mathrm{b}=14,\mathrm{c}=15$

Semi-perimeter$\left(\mathrm{s}\right)=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$

$$\begin{gathered} =\frac{13+14+15}{2} \\ =\frac{42}{2} \end{gathered}$$

$\therefore \mathrm{s}=21\mathrm{m}$

Step2:Find the area of triangle.

By using Heron's formula,

$\begin{array}{rcl}\mathrm{Area}\mathrm{of}\mathrm{triangle}& =& \sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}\\ & =& \sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}\\ & =& \sqrt{21\times 8\times 7\times 6}\\ & =& \sqrt{7\times 3\times 4\times 2\times 7\times 3\times 2}\\ & =& \sqrt{7\times 7\times 3\times 3\times 4\times 2\times 2}\\ & =& 7\times 3\times 2\times 2\\ & =& 84\end{array}$

$\therefore$Area of triangle is$84{\mathrm{m}}^2$.

Step3: Find the cost of advertisements for $6$ months .

Cost of advertisements for a year $=$ Area of triangle $\times$ Cost per ${\mathrm{m}}^2$

$$\begin{gathered} =84\times 2000 \\ =₹1,68,000 \end{gathered}$$

Cost of advertisements for $6$ months $=\frac{6}{12}\times 168000$

$=₹84000$

Hence, the cost of advertisements for $6$ months$=₹84000$