Question 1134255
<br>
The two widths are equal, and the two lengths are equal:
y = x^2
4y+8 = 20x-16  -->  4y = 20x-24  -->  y = 5x-6<br>
So<br>
x^2 = 5x-6
x^2-5x+6 = 0
(x-3)(x-2) = 0<br>
x could be either 3 or 2; check to see which (if either) satisfies all the conditions of the problem.<br>
If x = 2 then y = x^2 = 4; the dimensions are 2 and 4.  Not right -- the problem says one dimension has to be an odd integer.<br>
If x = 3 then y = x^2 = 9; the dimensions are 3 and 9.  This is okay; both dimensions are odd integers.<br>
Now you can find the perimeter to answer the question.