Question 1011123
{{{x}}}= side length of the smaller square, in cm,
{{{y}}}= side length of the larger square, in cm,
and {{{x<y}}} , of course.
So,
{{{x^2}}}= area of the smaller square, in square cm;
{{{y^2}}}= original area of the larger square, in square cm, and
{{{y^2-x^2}}}= remaining area of the larger square, in square cm.
The problem  tells us that {{{y^2-x^2=57}}} ,
and that {{{x}}} and {{{y}}} are positive integers less than 25.
Algebra tells us that
{{{y^2-x^2=57}}}<-->{{{(y-x)(y+x)=57}}} .
Since {{{x}}} and {{{y}}} are positive integers less than {{{25}}} , with {{{x<y}},
{{{y-x}}} and {{{y+x}}} are positive integers, and
{{{y-x<y+x<25+25=50}}} .
What positive integers less than {{{5}}} could be {{{y-x}}} and {{{y+x}}} ?
As {{{57}}} does not have too many factors, we can only write it as two products of positive integers:
{{{57=1*57}}} and {{{57=3*19}}} .
Only one of thpose products has both factors less than {{{50}}} .
So, {{{system(y-x=3,y+x=19)}}}---->{{{system(x=8,y=11)}}} .
The perimeter of the small square in centimetres is
{{{4*x=4*8=highlight(32)}}} .