What is the dimensions of the original square?
2007-08-31 15:03:44 · 6 answers · asked by Red 1 in Science & Mathematics ➔ Mathematics
A1 = x²
A2 = (x - 3)² = x² - 6x + 9
A1 - A2 = 81
6x - 9 = 81
6x = 90
x = 15
A1 is 15 x 15 cm²
2007-08-31 21:44:43 · answer #1 · answered by Como 7 · 1⤊ 0⤋
let the side of the original sq be x
X ^2 - (x-3) ^2 = 81
x ^ 2 - ( x ^ 2 - 6x + 9) = 81
6x -9 = 81
6x = 90
x = 15
15 x 15 = 225
12 x 12 = 144
diff = 81
2007-08-31 15:14:39 · answer #2 · answered by norman 7 · 0⤊ 0⤋
Let the original dimensions be X
X^2 - (X-3)^2 = 81
6X = 90
X = 15 cm
2007-08-31 15:10:19 · answer #3 · answered by sarjan_sarge 3 · 1⤊ 0⤋
Let, area = x^2
x^2 - 81 = (x - 3)^2
x^2 - 81 = x^2 - 6x + 9
6x = 90
x = 15
original side length is 15 cm.
2007-08-31 15:11:38 · answer #4 · answered by Anonymous · 1⤊ 0⤋
(x-3)(x-3)=x^2 - 81
x^2 - 6x + 9 = x^2 - 81
-6x = -90
x = 15
2007-08-31 15:12:14 · answer #5 · answered by timemccormick 3 · 0⤊ 1⤋
x^2 -- 81 = (x -- 3)^2
=> 6x = 90
=> x = 15
side is 15cm
2007-08-31 15:14:06 · answer #6 · answered by sv 7 · 0⤊ 1⤋