Question 705910
Answer: Yes

Proof:
Let the Let the width and height be denoted by {{{w}}} and {{{h}}} respectively

{{{ perimeter = 2w+2h = 24 }}}
{{{ w+h = 12 }}}   Eq (1)


{{{ area = wh=36 }}}    Eg (2)


From Eq(1) we get {{{ w = 12-h }}}


Substituting into Eq (2) we get:
{{{ wh=36 }}}
{{{ (12-h)h=36 }}}
{{{ 12h-h^2=36 }}}
{{{ h^2-12h+36=0 }}}
{{{ (h-6)(h-6)=0 }}}
Giving the solution {{{h=6}}} (repeated)


And from Eq(1) when {{{h=6}}} then {{{w=6}}}


so a rectangle of length 6 (ie square) satisfies conditions.


The same process will work for other Numbers