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

Explanation

`a,b, and c`, be the sides of the orignial tirangle
Then`s=1/2(a+b+c)`
Area`(Δ)=sqrt(s(s-a)(s-b)(s-c)`
Let the sides of the new triangle`=2a,2b,2c`
`s=1/2(2a+2b+2c)`
`=2xx(a+b+c)/2`
`2s=(a+b+c)`
and area `Delta=sqrt(s(2s-2a)(2s-2b)(2s-2c)`
`=sqrt(s(2xx2xx2xx(s−a)(s−b)(s−c))`
`=4sqrt (s(s-a)(s-b)(s-c)`
`= 4Delta`
Hence area of new triangle=4(area of original triangle).