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 (see Fig. 129). 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?
122md
22m
120m
Fig. 12.9​

Answer 1

Given :-

Sides of the triangle ABC = 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.

To Find :-

The area of the triangle.

The rent of one wall for 3 months.

Analysis :-

At first, find the perimeter and the semi perimeter of the triangle.

Then using the Heron's formula, find the area of the triangle.

Finally, get the cost accordingly by multiplying.

Solution :-

We know that,

• s = Side
• a = Area

By the formula,

$\underline{\boxed{\sf Semi \ perimeter=\dfrac{a+b+c}{2} }}$

Given that,

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

Substituting their values,

$\sf Semi \ perimeter=\dfrac{122+22+120}{2}$

$\sf Semi \ perimeter = \dfrac{264}{2} =132 \ m$

Therefore, the semi perimeter is 132 m

Using Heron’s formula,

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

Substituting their values,

$\sf =\sqrt{132(132-122)(132-22)(132-120)}$

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

$\sf =1320 \ m^{2}$

Therefore, the area of triangle is 1320 m²

Given that,

Rent of advertising per year = Rs. 5000 per m²

The rent of one wall for 3 months,

$\sf \dfrac{1320 \times 5000 \times 3}{12} =1650000$

Therefore, the cost of one wall for 3 months is Rs. 1650000

Answer 2

Answer:

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