Let Δ be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.

#### Solution

We are given assumed value Δ is the area of a given triangle *ABC*

We assume the sides of the given triangle *ABC* be *a,* *b,* c

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

`A = sqrt(s(s-a)(s-b)(s-c))`

`Δ = sqrt( s(s-a)(s-b)(s-c))` ...................(1)

Where,

`s =(a+b+c)/2`

`2s = a +b+c` ....................(2)

We take the sides of a new triangle as 2*a*, 2b, 2c that is twice the sides of previous one

Now, the area of a triangle having sides 2*a**,* 2*b**,* and 2c* *and s_{1} as semi-perimeter is given by,

`A_1 = sqrt(s_1(s_1-2a)(s_1-2b)(s_1-2c))`, where

`s_1 = ( 2a+2b+2c)/2`

`s_1 = (2(a+b+c))/2`

s_{1 }= a+ b+ c

s_{1 }=2s ..................(using (2))

Now,

`A_1 = sqrt(2s(2s-2a)(2s-2b)(2s-2c)`

`A_1= sqrt(2x xx 2 (s-a) xx 2 (s-b) xx 2(s-c))`

`A_1 = 4sqrt(4(s-a)(s-b)(s-c))`

`A_1 = 4Δ` ...................(using (1))