Question

the triangular side walls of a flyover have been used for advertisement. the side of the walls are 122m,22m and 120m . the advertisement yield on earning of ₹ 5000 Per m square per year. A company hired one of its walls for 3 months . how much rent did it pay?​

Answer 1

Answer:

Rs 16,500,000

Step-by-step explanation:

Refer figure,

In △ABC

a=122m,b=22m,c=120m

s=  

2

a+b+c

​  

=  

2

122+22+120

​  

 

=132m

ar△ABC=  

s(s−a)(s−b)(s−c)

​  

=  

132(132−122)(132−23)(132−120)

​  

 

=  

132×10×110×12

​  

=  

132×10×11×10×12

​  

=  

(132)  

2

×(10)  

2

 

​  

 

=132×10=1320 m  

2

 

Rent of 1 m  

2

 per year = Rs. 5000

Rent of 1 m  

2

 per month = Rs.  

12

5000

​  

 

Rent of 1 m  

2

 for 3 months=Rs.  

12

3×5000

​  

 

Rent of 1320 m  

2

 for 3 months=Rs. (  

12

3×5000×1320

​  

)

= Rs.16,500,000

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