Question

Triangular side wall of flyover have been for advertisements. the sides of the walls are 122 m and 22 m and 120 m. the advertisements yield and earning of rupees 5000 per metre square year. a company hired one of its wall for 3 month. how much rent did it pay?​

Answer 1

The perimeter of a triangle is equal to the sum of its three sides it is denoted by 2S.

2s=(a+b+c)

s=(a+b+c)/2

Here ,s is called semi perimeter of a triangle.

The formula given by Heron about the area of a triangle is known as Heron's formula.

According to this formula area of a triangle= √s (s-a) (s-b) (s-c)

Where a, b and c are three sides of a triangle and s is a semi perimeter.

This formula can be used for any triangle to calculate its area and it is very useful when it is not possible to find the height of the triangle easily .

Heron's formula is generally used for calculating area of scalene triangle.

___________________ _______________

Solution:

Let the sides of the triangle are a=122 m, b=22 m & c= 120 m.

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

s=(122 + 22 + 120) / 2

s= 264/2= 132m

Using heron’s formula,

Area of the wall = √s (s-a) (s-b) (s-c)

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

= √132 × 10 × 110 × 12

=√11×12×10×11×10×12

=√11×11×12×12×10×10

= 11×12×10

= 1320m²

Given, earning on 1m² per year= ₹5000

Earning on 1320 m² per year=1320×5000= ₹6600000

Now, earning in 1320 m² in 12 months= ₹6600000

earning in 3 months = ₹ 6600000 ×3/12 = ₹ 1650000

Hence, the rent paid by the company for 3 months is ₹ 1650000.

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Answer 2

$\mathfrak{\huge{\pink{\underline{Given:-}}}}$

➤ The sides of the triangle ABC = 122 m, 22 m and 120 m respectively

➤ Rent of advertising per year = ₹ 5000 per m²

$\mathfrak{\huge{\pink{\underline{To \: Find:-}}}}$

➤ The rent of one wall for 3 months.

$\mathfrak{\huge{\pink{\underline{Solution:-}}}}$

Given that,

The sides of the triangle ABC are 122 m, 22 m and 120 m respectively.

Now,

The perimeter $\sf=122+22+120$

$\sf =264 \: m$

Also,

Semi perimeter (s) $\sf =\dfrac{264}{2}$

$\sf =132 \: m$

Using Heron’s formula,

Area of the triangle,

$\sf \sqrt{s(s-a)(s-b)(s-c)}$

$\sf \longrightarrow \sqrt{132(132-122)(132-22)(132-120)} \: m^{2}$

$\sf \longrightarrow \sqrt{132 \times 10 \times 110 \times 12} \: m^{2}$

$\sf \longrightarrow 1320 \: m^{2}$

We know that,

Rent of advertising per year = ₹ 5000 per m²

∴ The rent of one wall for 3 months = $\sf Rs. \: \dfrac{1320 \times 5000 \times 3}{12}$

$\implies \sf \underline{\underline{Rs. \: 1650000}}$