Question 13 Marks
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $13 m, 14 m$ and $15 m $. The advertisements yield an earning of Rs. $2000 $per $m ^2$ a year. A company hired one of its walls for $6$ months. How much rent did it pay?
Answer
View full question & answer→Since, the sides of a triangular walls are $a=13 m, b =14 m$ and $c =15 m$
$\therefore$ Semi-perimeter of triangular side wall, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{13+14+15}{2}=\frac{42}{2}=21\text{m}$
$\therefore$ Area of triangular side wall, $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula] $=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{21\times8\times7\times6}$
$=\sqrt{21\times4\times2\times7\times3\times2}$
$=\sqrt{(21)^2\times(4)^2}$
$=21\times4=84\text{m}^2$
Since, the advertisement yield eaming per year for $1 m^2=$ Rs. $2000 $
$\therefore$ Advertisement yield earning per year on $84 m^2=2000 \times 84=$ Rs. 168000 As the company hired one of its walls for $6$ moths, therefore company pay the rent $=\frac{1}{2}( 1 6 8 0 0 0 )=$ Rs. 84000 Hence, the company6 paid tent Rs. $84000$
$\therefore$ Semi-perimeter of triangular side wall, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{13+14+15}{2}=\frac{42}{2}=21\text{m}$
$\therefore$ Area of triangular side wall, $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula] $=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{21\times8\times7\times6}$
$=\sqrt{21\times4\times2\times7\times3\times2}$
$=\sqrt{(21)^2\times(4)^2}$
$=21\times4=84\text{m}^2$
Since, the advertisement yield eaming per year for $1 m^2=$ Rs. $2000 $
$\therefore$ Advertisement yield earning per year on $84 m^2=2000 \times 84=$ Rs. 168000 As the company hired one of its walls for $6$ moths, therefore company pay the rent $=\frac{1}{2}( 1 6 8 0 0 0 )=$ Rs. 84000 Hence, the company6 paid tent Rs. $84000$