If each side of a triangle is doubled then find the ratio of the area of the new triangle thus formed and the given triangle.

Let sides of the triangle be a, b and c.

Then, semi-perimeter of a triangle is given as

…(i)

And so, the area of a triangle is given as

Area of triangle = √(s(s – a)(s – b)(s – c)) …(ii)

According to the question,

Each side of the triangle is doubled.

So, sides of new triangle are a’ = 2a, b’ = 2b and c’ = 2c.

Then, semi-perimeter of the new triangle will be given as

⇒ s’ = a + b + c …(iii)

Comparing equations (i) and (iii), we get

⇒ s’ = 2s …(iv)

And thus, area of new triangle is given as

Area of new triangle = √(s’(s’ – a’)(s’ – b’)(s’ – c’))

Substitute s’ = 2s from equation (iv) & a’ = 2a, b’ = 2b and c’ = 2c in the above equation, we get

Area of new triangle = √(2s(2s – 2a)(2s – 2b)(2s – 2c))

⇒ Area of new triangle = √((2 × 2 × 2 × 2)s(s – a)(s – b)(s – c))

⇒ Area of new triangle = 4√(s(s – a)(s – b)(s – c))

⇒ Area of new triangle = 4 × Area of triangle

As we know,

Area of new triangle = 4 × Area of triangle

Thus, the ratio of Area of the new triangle to the Area of a triangle is 4:1.