• June 28th 2007, 02:13 PM
beautiful
if the sides of a square are decreased by 2 cm, the area is decreased by 36cm^2, what were the dimensions of the original square.

im thinking that dimension is the sides

a=s^2 right? i forget the formulas.

sides = x-2

area= x-36cm^2

if this is right so far can you help me from here or correct what i have and explain. thanks

:D
• June 28th 2007, 02:29 PM
galactus
You have the original square: $A=x^{2}$

decrease the sides by 2 and the area decreases by 36:

$(x-2)^{2}=A-36$

From the first equation $x=\sqrt{A}$

Then we have:

$(\sqrt{A}-2)^{2}=A-36$

$A-4\sqrt{A}+4=A-36$

$A-4\sqrt{A}=A-40$

$\sqrt{A}=10$

$A=100$