Question

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?

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Answer 1

Answer:

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.

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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

Given :

• Sides of triangle = 122 m , 22 m , 120 m

• Cost of advertisement = ₹ 5000 per m² per year

• Time = 3 months = 1/4 year

To Find :

• Cost of advertisement for 1/4 year

Solution :

Firstly we have to find Semi-perimeter of wall

$\large\longrightarrow \boxed{\boxed{ \green{\sf s = \dfrac{a + b + c}{2}}}} \\ \\ \sf \longrightarrow s = \frac{122 + 22 + 120}{2} \\ \\\sf \longrightarrow s = \frac{264}{2} \\ \\ \longrightarrow \boxed{\sf s = 132}$

Now by using Heron's formula we will find Area of wall

$\large \longrightarrow \boxed {\boxed{ \red{ \sf Area = \sqrt{s(s - a)(s -b)(s-c)}}}}\\ \\\sf \longrightarrow \sqrt{132(132 - 122)(132 - 22)(132 - 120)} \\ \\\sf \longrightarrow \sqrt{132 \times 10 \times 110 \times 12} \\ \\\sf \longrightarrow \sqrt{ 12 \times 11 \times 10 \times 11 \times 10 \times 12} \\ \\\sf \longrightarrow 12 \times 11 \times 10 \\ \\\large\longrightarrow \boxed{ \sf Area = 1320 \: {m}^{2}}$

Now Cost of Advertisement at rate of 5000 per m² for 1/4 year

$\sf \longrightarrow Cost = 1320 \times 5000 \times \frac{1}{4} \\ \\\sf \longrightarrow Cost = 330 \times 5000 \\ \\\large \longrightarrow \boxed{\sf \orange{ Cost = 16,50,000 \: Rs}}$