Questions

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.). 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?

detailed solution

Correct option is A

We know that,

The Heron's formula for calculating the area of the triangle,

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

where, s is the semi-perimeter = half of the perimeter

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

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

Thus, Semi-perimeter

s = (a $+$ b $+$ c)/2

= (122 $+$ 22 $+$ 120)/2

= 264/2

= 132 m

Area of a triangular wall = $\sqrt{132(132-122)(132-22)(132-120)}$

= $\sqrt{132\times 10\times 110\times 12}$

= 1320 m

Given that,

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) $\times $ 3 $\times $1320

= ₹ 1650000

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

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detailed solution

Correct answer is 1

We know that,

The Heron's formula for calculating the area of the triangle,

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

where, s is the semi-perimeter = half of the perimeter

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

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

Thus, Semi-perimeter

s = (a $+$ b $+$ c)/2

= (122 $+$ 22 $+$ 120)/2

= 264/2

= 132 m

Area of a triangular wall = $\sqrt{132(132-122)(132-22)(132-120)}$

= $\sqrt{132\times 10\times 110\times 12}$

= 1320 m

Given that,

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) $\times $ 3 $\times $1320

= ₹ 1650000

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

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