Question

the triangular side walls are 122m 22m and 120m the advertisement yield an earning of rs 5000 per m2 per year a company hired one of its walls for 3 months how much rent did it pay?​

Answer 1

Given :-

Side A of the triangle = 122 m

Side B of the triangle = 22 m

Side C of the triangle = 120 m

The advertisements yield an earning per year =  Rs. 5000 per m²

To Find :-

The area of the triangle.

The rent of one wall for 3 months.

Analysis :-

Get the perimeter and the semi perimeter of the triangle.

Next the area of the triangle using the heron's formula.

Multiply the area of the triangle by the rent advertising per year.

Solution :-

Given that,

Side A = 122 m

Side B = 22 m

Side C = 120 m

Finding the perimeter,

$\underline{\boxed{\sf Perimeter \ of \ a \ triangle= A+B+C}}$

Perimeter of the triangle = $\sf 122+22+120$

$\longrightarrow \sf 263 \ m$

Then the semi-perimeter would be,

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

Using Heron's formula,

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

Substituting their values,

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

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

$\sf =\sqrt{1742400}$

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

Given that,

The rent of advertising per year = Rs. 5000 per m²

∴ The rent of one wall for 3 months,

$\sf =Rs. \ \dfrac{1320 \times 5000 \times 3}{2}$

$\sf =Rs. \ 1650000$

Therefore, he paid Rs. 1650000

Answer 2

Given :-

• The triangular side walls are 122m 22m and 120m .

• The advertisement yield an earning of rs 5000 per m2 per year .

• A company hired one of its walls for 3 months

To find :-

• how much rent did it pay?

NOTE :-

To find how much he paid, we must first find

• Area of the triangle &

• Rent of walls for 3 months .

Solution :-

We know ,

$\underline{\boxed{\sf \pink{✷Perimeter \ of \ a \ triangle= A+B+C}}}$

Substituting the values,

→ Perimeter = 122+22+120

$\implies \sf 263 \ m$

Then the semi-perimeter would be,

$\implies \sf \dfrac{264}{2}=132 \ m$

Now,

Heron's formula :

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

$✧\underline{\purple{\sf S = a+b+c/2 }}$

Substituting their values,

$\sf =\sqrt{132(132-122)(132-22)(132-120)}\\\\\\</p><p> \sf =\sqrt{132 \times 10 \times 110 \times 12}\\\\\\</p><p> \sf =\sqrt{1742400}\\\\\\</p><p>\sf =1320 \ m^{2}$

And,

The rent of one wall for 3 months :

$\sf =Rs. \ \dfrac{1320 \times 5000 \times 3}{2}\\\\\\</p><p>\sf =Rs. \ 1650000$

Hence, $\pink{\underline{\boxed {He \: paid \: Rs.1650000}}}$

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