2007-01-30 08:02:06 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
This can be solved by simply applying the difference of the two squares. Say the original square has sides of length t, thus the second square would have sides of length (t-2).
That taken into consideration t^2 - (t-2)^2 = 36.
The difference of the two squares was 36 cm^2.
note: (t-2)^2 = t^2 - 4t + 4
When plugged into the above area equation we get:
t^2 - t^2 +4t - 4 = 36. The (t^2)s cancel out leaving 4t = 40.
Thus t=10.
The first square had sides of 10 cm and the second square had sides of 8 cm.
2007-01-30 08:08:22 · answer #1 · answered by uahgrad05 3 · 2⤊ 0⤋
If you call t the length of the side
A = t^2 and A-36 =(t-2)^2
t^2 -36 = t^2-4t +4 so 4t =40 and t =10cm
2007-01-30 20:28:04 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
- 2 cm for each side..right? than the surface is -(2cm*2cm)=-4cm^2....
2007-01-30 16:08:59 · answer #3 · answered by ddroxana 2 · 0⤊ 0⤋