Suppose that, in #DeltaABC, BC=a, CA=b, AB=c#.
Also, let #[ABC]# denote the Area of #DeltaABC#.
Then, by Heron's Formula,
#[ABC]=sqrt(s(s-a)(s-b)(s-c)); s=(a+b+c)/2#.
Consider #DeltaA'B'C'# in which,
#B'C'=a'=2BC=2a, C'A'=2b, and A'B'=2c#.
Hence, #s'=(a'+b'+c')/2=(2a+2b+2c)/2=a+b+c=2s#.
#s'-a'=2s-2a=2(s-a)#.
Similarly, #s'-b'=2(s-b), and s'-c'=2(s-c)#.
#:.[A'B'C']=sqrt(s'(s'-a')(s'-b')(s'-c'))#,
#=sqrt(2s*2(s-a)*2(s-b)*2(s-c))#,
#=4sqrt(s(s-a)(s-b)(s-c))#.
#rArr [A'B'C']=4[ABC], or #
#x_2=4x_1#.
#:." The increase in the area="x_2-x_1=3x_1#.
#:." The % increase in the area="(x_2-x_1)/x_1*100#,
#=(3x_1)/x_1*100=300%#,