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?

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- Jun 30th 2007, 10:31 AMlizard4geometry help
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? - Jun 30th 2007, 10:40 AMJhevon
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 - Jun 30th 2007, 10:40 AMIlaggoodly
why seperate the length and the width, its a square so take advantage of the fact that all the sides are the same length.

In this scenario you have two equations,

s^2 = A (the original square)

(s-2)^2 = A-36 (the new square)