# geometry help

Printable View

• Jun 30th 2007, 10:31 AM
lizard4
geometry 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 AM
Jhevon
Quote:

Originally Posted by lizard4
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
• Jun 30th 2007, 10:40 AM
Ilaggoodly
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)