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


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Answer 1

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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 m² area per year = ₹ 5000

Rent of 1 m² area per month = ₹ 5000/12

Rent of 1320m² area for 3 months

= ₹ (5000/12) × 3 × 1320

= ₹ 1650000

Therefore,

The company paid ₹ 16,50,000 as rent.

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Answer 2

According to the figure : [ In the attachment ]

Sides of a triangular wall is:

a = 122 m

b = 22 m

c = 120 m

And also given:

The advertisements yeild an earning

of 5000/m² per year.

We have to find :

The rent paid by the company for 3 months

Solution:

$\begin{array}{cc}\sf \: semi \: perimeter \: \: s = \frac{a + b + c}{2} \\ \\ \sf = \frac{122 + 22 + 120}{2} \\ \\ = \sf \: 132 \: m \end{array}$

Area of the triangle :

${{\boxed{\begin{array}{cc} \sf \: \Delta \: = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{132 \times (132 - 122)(132 - 22)(132 - 120)} \\ \\ = \sqrt{132 \times 10 \times 110 \times 12} \\ \\ = 1310 \: {m}^{2} \end{array}}}}$

Now,

Earning on 1320 m² per year

=1320×5000= 6600000

So, earning in 3 months = 6600000 × 3/12 = 1650000

Finally, the rent paid by the company for 3 months is = 1650000.