Question 862466
The length of a rectangle is k inches more than its width.  If perimeter p is known, what is area A?


Let L = length
Let w = width
k is a described constant, as is p.
A is a variable, unknown, but calculable, for the area.
-
{{{L=w+k}}}.


{{{2w+2L=p}}}
{{{2w+2(w+k)=p}}}
{{{4w+2k=p}}}
{{{4w=p-2k}}}
{{{highlight(w=(p-2k)/4)}}}
Use w to find L.
{{{L=w+k}}}
{{{L=(p-2k)/4+k}}}------this form may be good enough...
{{{L=(p-2k)/4+4k/4}}}
{{{L=((p-2k)+4k)/4}}}
{{{highlight(L=(p+2k)/4)}}}  -------this form seems better.


Now, you want Area A, and you begin forming {{{highlight_green(A=w*L)}}}.
{{{A=((p-2k)/4)((p+2k)/4)}}}
{{{highlight(highlight(A=(p^2-(4k^2))/16))}}}


The solution is generalized.  You should now appreciate the use of purely solving all in symbols.  You have a formula for the area which was asked.  What would be important is that the values for p and k make sense in the described situation.  Their actual values are not too important; only that they make sense in the situation is important.  In your example, p=36 and k=2.  Use them and compute A.