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CBSE Class 9 Mathematics Question 2 of 6

Heron's Formula — Question 2

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Question 2

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 m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

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<sup>2</sup> per year. A company hired one of its walls for 3 months. How much rent did it pay? NCERT Class 9 Mathematics CBSE Solutions.
Answer

Here a, b and c are the sides of the triangle.

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

By formula,

Semi Perimeter (s) = Perimeter of triangle2\dfrac{\text{Perimeter of triangle}}{2}

s = a+b+c2=122+22+1202=2642\dfrac{a + b + c}{2} = \dfrac{122 + 22 + 120}{2} = \dfrac{264}{2} = 132 m.

By Heron's formula,

Area of triangle (A) = s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)} sq.units

Substituting values we get :

Area of one wall=132(132−122)(132−22)(132−120)=132×10×110×12=1742400=1320 m2.\text{Area of one wall} = \sqrt{132(132 - 122)(132 - 22)(132 - 120)} \\[1em] = \sqrt{132 \times 10 \times 110 \times 12} \\[1em] = \sqrt{1742400} \\[1em] = 1320 \text{ m}^2.

We know that,

The rent of advertising per year = ₹ 5000 per m2

So,

The rent of one complete triangular wall for 1 month

= Rent per sq. unit× Area12\dfrac{\text{Rent per sq. unit} \times \text{ Area}}{12}

= (5000×1320)12=11×5000\dfrac{(5000 \times 1320)}{12} = 11 \times 5000 = ₹ 5,50,000.

∴ The rent of one wall for 3 months = ₹ 5,50,000 x 3 = ₹ 16,50,000.

Hence, the rent paid by the company = ₹ 16,50,000.