Question 1030634
Size is in reference to area, not distances.


4 by 6 is area  {{{4*6=24*inch^2}}}.


{{{4*6=(2/7)A}}} using original area A.


{{{A=(4*6)(7/2)}}}
{{{A=2*6*7}}}
{{{A=84}}}


Some linear factor k will change 4 by 6 into the area 84.
{{{4k*6k=84}}}
{{{4*6*k^2=84}}}
{{{k^2=84/(4*6)}}}
{{{k^2=(7*2*6)/(4*6)}}}
{{{k^2=(7*2)/4}}}
{{{k^2=7/2}}}
{{{k=sqrt(7/2)}}}
{{{k=sqrt(7)sqrt(2)/2}}}
{{{k=(1/2)sqrt(14)}}}


Original dimensions were {{{highlight(4(1/2)sqrt(14)=2sqrt(14))}}}
and
{{{highlight(6(1/2)sqrt(14)=3sqrt(14))}}}.