Question

The triangular side wall of flyover have been used for advertisment.the sides of the walls are 122m,22m and 120m.the advertisment yeild an earning of rs.5000 per m² per year.a company hired one of its walls for three months .how much rent did he pay.​

Answer 1

Answer:

Step-by-step explanation:

using heron's formula

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

s=a+b+c/2= 122+22+120/2 = 264/2= 132

AREA= √s(s-a)(s-b)(s-c)= √132(132-122)(132-22)(132-120)

        =√132(10)(110)(12)= √1742400= 1320m∧2

Let 'y' be rupees be the rent per month

then,according to the question,

rent per year= 12 × 1320 × y =5000rupees    (1 yr=12 months)

                        y = 5000/1320×12 rupees

rent for 3 months = 3 × 1320 × y

                             =3 × 1320× 5000/1320×12 =5000/4 = 1250 rupees

hope this helps.

Answer 2

Given:-

• the sides of the walls are 122m,22m and 120m.
• The advertisment yeild an earning of rs.5000 per m² per year.
• A company hired one of its walls for three months .

To find:-

• Find the company rent did he pay.

Solutions:-

• The side of the triangle are of 122m, 22m and 120m.

perimeter of triangle = (a + b + c)

= (122 + 22 + 120)m

= 264m

S = (perimeter of triangle)/2

S = 264/2

S = 132m

By heroe's formula,

Area of triangle = √s(s - a)(s - b)(s -c)

= √132(132 - 122)(132 - 22)(132 - 120)

= √132 × 10 × 110 × 12

= √1742400

= 1320m²

Now,

Rent of 1 m² area per year = Rs 5000

Rent of 1 m² area per month = Rs 5000/2

Rent of 1320 m² area for 3 month

= Rs (5000/12 × 3 × 1320)

= Rs 5000 × 330

= Rs 1650000

Hence, the company had to pay Rs 1650000.