# The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m, and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹ 5000 per m^{2} per year. A company hired one of its walls for 3 months. How much rent did it pay?

**Solution:**

Given: Dimensions of the triangular sides of walls.

By using Heron’s formula, we can calculate the area of triangle.

Heron's formula for the area of a triangle is: Area = √s(s - a)(s - b)(s - c)

Where a, b and c are the sides of the triangle, and s = Semi-perimeter = Half the Perimeter of the triangle

Triangular sides of walls are, a = 122 m, b = 22 m, c = 120 m

Semi Perimeter, s = (a + b + c)/2

= (122 + 22 + 120)/2

= 264/2

= 132 m

By using Heron’s formula,

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

Substuting the values in order to find area of triangular wall,

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

= √132 × 10 × 110 × 12

= 1320 m

Rent of 1 m^{2} area per year = ₹ 5000

Rent of 1 m^{2} area per month = ₹ 5000/12

Rent of 1320 m^{2 }area for 3 months = ₹ (5000/12) × 3 × 1320

= ₹ 1650000

Therefore, the company paid ₹ 16,50,000 as rent.

**Video Solution:**

## The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m, and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹ 5000 per sq. m per year. A company hired one of its walls for 3 months. How much rent did it pay?

### Class 9 Maths NCERT Solutions - Chapter 12 Exercise 12.1 Question 2:

**Summary:**

It is given that the triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m, and 120 m. The advertisements yield an earning of ₹ 5000 per m^{2} per year. A company hired one of its walls for 3 months. Therefore, the company paid ₹ 16,50,000 as rent.