What will be the increase in …

Let a,b,c be the sides of the original ∆ & s be its semi perimeter.
S= (a+b+c)/2
2s= a+b+c.................(1)
The sides of a new ∆ are 2a,2b,2c [ given: Side is doubled]

Let s' be the new semi perimeter.
s'= (2a+2b+2c)/2
s'= 2(a+b+c) /2
s'= a+b+c
S'= 2s. ( From eq 1)......(2)
Let ∆= area of original triangle
∆= √s(s-a)(s-b)(s-c).........(3)

&

∆'= area of new Triangle
∆' = √s'(s'-2a)(s'-2b)(s'-2c)
∆'= √ 2s(2s-2a)(2s-2b)(2s-2c) [From eq. 2]
∆'= √ 2s×2(s-a)×2(s-b)×2(s-c)
= √16s(s-a)(s-b)(s-c)
∆'= 4 √s(s-a)(s-b)(s-c)
∆'= 4∆. (From eq (3))
Increase in the area of the triangle= ∆'- ∆= 4∆ - 1∆= 3∆
%increase in area= (increase in the area of the triangle/ original area of the triangle)× 100
% increase in area= (3∆/∆)×100
% increase in area= 3×100=300 %
Hence, the percentage increase in the area of a triangle is 300%.