Question

The triangular side's walls of a flyover have been used for advertisements. The sides of the
walls are 122m. 22m and 120m. The advertisement yield on earning of Rs. 5000 per per
year. A company hired one of its walls for 4 months. How much rent did it pay?​

Answer 1

In ΔABC:

a=122m

b=120m

c=22m

b

2

+c

2

=120

2

+22

2

=14884=122

2

=a

2

b

2

+c

2

=a

2

∴ΔABC is Right angle triangle

AreaΔABC(σ)=

2

1

×120×22=1320m

2

Cost of Advertisement C=Rs.5000/m

2

/Year

Time for which wall is hire t=3months=

4

1

year

area occupied σ=1320m

2

Total cost C=5000×1320×

4

1

∴C=Rs.1,650,000

Answer 2

Given:-

• The triangular side's walls of a flyover have been used for advertisements. The sides of thewalls are 122m. 22m and 120m.

• The advertisement yield on earning of Rs. 5000 per per year.

To Find:-

• How much rent did it pay in 4 months.

Formulae used:-

• Heron's Formulae

Now,

• a = 122m

• b = 22m

• c = 120m

Therefore,

→ $\sf{ S = \dfrac{Perimeter}{2}}$

→ $\sf{ S = \dfrac{a + b + c}{2}}$

→ $\sf{ S = \dfrac{122 + 22 + 120}{2}}$

→ $\sf{ S = \dfrac{\cancel{262}}{\cancel{2}}}$

→ $\sf{S = 132}$.

Now, Using Heron's Formulae

→ Area of ∆ABC = $\sf{\sqrt{(s) ( s - a ) ( s - b ) ( s - c )}}$

→ Area of ∆ABC = $\sf{\sqrt{( 132) ( 132 - 122 ) ( 132 - 22 ) ( 132 - 120 )}}$

→ $\sf{\sqrt{(132) ( 10 ) ( 110 ) ( 12 )}}$

→ $\sf{\sqrt{1742400}}$

→ $\sf{1320}$

Hence, The Area of Triangle is 1320m².

Now,

→ Cost of Advertising Per m² = $5000

→ Cost of Advertising 1320m² = $5000 × 1320

→ Cost of Advertising →$6600000

Now, Atq

→ $\sf{Cost\:of\:Advertising\:it's\:one\:wall\:for\: four\:months = 6600000\times{\dfrac{3}{12}}}$

→ $\sf{ 6600000 ÷ 4 }$

→ $\sf{ 1650000}$.

Hence, The rent he would pay is $1650000