Question

ਇੱਕ ਤ੍ਰਿਭੁਜ ਆਕਾਰ ਦੇ ਭੂ-ਖੰਡ ਦੀਆਂ ਭੁਜਾਵਾਂ ਦਾ ਅਨੁਪਾਤ 3:5:7 ਹੈ।ਇਸ ਦਾ ਪਰਿਮਾਪ 300 ਮੀਟਰ ਹੈ। ਇਸ ਭੂ-ਖੰਡ ਦਾ ਖੇਤਰਫਲ ਕਿੰਨਾ ਹੋਵੇਗਾ ​

Answer 1

Given:

A triangle-shaped land has the sides  in the ratio 3:5:7

Its perimeter is  300 meter

To find:

The area of  the land

Solution:

Let's assume "3x", "5x" & "7x" are the 3 sides of the triangular-shaped land.

We know,

The perimeter of a triangle = sum of all sides of the triangle

But we have,

The perimeter of land = 300 m

⇒ $3x + 5x + 7x = 300$

⇒ $15x = 300$

⇒ $x =$ $\frac{300}{15}$

⇒ $\bold{x = 20}$

Therefore,  

The first side of the triangle = 3x = 3 × 20 = 60 m

The second side of the triangle = 5x = 5 × 20 = 100 m

The third side of the triangle = 7x = 7 × 20 = 140 m

To find the area of the triangular-shaped land we will use the Heron's formula which is as follows:

If 3 sides of the triangle are a, b & c and the semi-perimeter is represented as "S" then,

$\boxed{\bold{Semi-perimeter, S = \frac{a+b+c}{2} }}\\\\\boxed{\bold{Area = \sqrt{S(S-a)(S-b)(S-c)} }}$

Now,

The semi-perimeter of land, S = $\frac{60+ 100+ 140}{2} = \frac{300}{2} = 150 \:m$

Substituting the values of S and a, b & c in the formula of Area, we get

$Area = \sqrt{150 (150-60)(150-100)(150-140)}$

$\implies Area = \sqrt{150 \times 90\times 50\times 10$

$\implies Area = \sqrt{3 \times 50 \times 3\times 3\times 10 \times 50 \times 10}$

$\implies Area = \sqrt{3^2 \times 50^2 \times 3\times 10^2}$

$\implies Area = 3 \times 50 \times 10\sqrt{ 3}$

$\implies \bold{Area = 1500\sqrt{ 3}\:m^2}$

Thus, the area of the triangular-shaped land is → $\underline{\bold{1500\sqrt{3} \:m^2}}$.

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