If the sides of a square are decreased by 2cm, the area is decreased by 36m^2. What are the dimensions of the original square?
So A(x-36cm^2)=Lenght 2(x-2cm + Width 2(x-2cm) Is this the right set up?
Let $\displaystyle x$ be the side length of the original square
Let $\displaystyle A$ be the area of the original square
Then we have: $\displaystyle x^2 = A$ ............(1)
When the side-length is decreased by 2, the area decreases by 36, so we also have:
$\displaystyle (x - 2)^2 = A - 36$
$\displaystyle \Rightarrow x^2 - 4x + 4 = A - 36$
$\displaystyle \Rightarrow x^2 - 4x + 40 = A$ ..............(2)
Now equate both formulas for A and solve for x, we get:
$\displaystyle x^2 - 4x + 40 = x^2$
And i think you can take it from there