Question 892432
A=1131
p=136

Let dimensions be x and y.


LIST WHAT YOU KNOW INCLUDING INFORMATION FROM THE DESCRIPTION
p=2x+2y;
A=xy.


Solve for x and y.  A quadratic equation will occur.  Do everything in variables until x and y are solved, AS FORMULAS, ALL IN VARIABLES.  


Substitute the values for A and p and evaluate x and y.



-----The less general way-------

2x+2y=136 and xy=1131.
Perimeter simplifies to {{{x+y=68}}} from which {{{y=68-x}}}.
-
{{{xy=1131}}}
{{{x(68-x)=1131}}}
{{{68x-x^2=1131}}}
{{{68x-x^2-1131=0}}}
{{{x^2-68x+1131=0}}}
{{{x=(68+- sqrt(68^2-4*1131))/2}}}
{{{x=(68+- sqrt(100))/2}}}
{{{x=(68+- 10)/2}}}
{{{x=29}}} or {{{x=39}}}


Checking for corresponding y values, you'll find in respective order, 39 and 29.
ANSWER: <b>Dimensions are 29 and 39</b>.