If the sides of a square are decreased by 3 cm, the area is decreased by 81 cm2. What were the dimensions of the original square?
2006-12-10 12:59:42 · 11 answers · asked by Deidrebm 1 in Science & Mathematics ➔ Mathematics
(x-3)^2 = area -81
x^2=area
subsitute
(x-3)^2 = x^2-81
x^2 - 6x + 9 = x ^2 - 81
-6x= -90
x=15
Orignal side was 15, new side 12
2006-12-10 13:10:30 · answer #1 · answered by bkc99xx 6 · 0⤊ 0⤋
Let X = the original side of the square and (X-3) be the length of the new side. Subtract the area of the new square from the area of the old square and set it to 81. You should be able to expand the equation and solve for X.
X^2 - (X-3)^2 = 81
2006-12-10 13:04:23 · answer #2 · answered by Gene 7 · 0⤊ 0⤋
4
2006-12-10 13:01:13 · answer #3 · answered by clair 1 · 0⤊ 0⤋
a - a' = 3 and a^2 - (a')^2 = 81
a + a' = 81 / 3 = 27
a + a' = 27 and a - a' = 3 => a = 15cm ; a' = 12cm
the initial side of the square is 15cm
2006-12-10 13:07:49 · answer #4 · answered by James Chan 4 · 0⤊ 0⤋
Original side = x. (x-3)^2 = x^2 - 81.
x = ?
2006-12-10 13:03:39 · answer #5 · answered by steve_geo1 7 · 0⤊ 0⤋
15cm;
15x15= 225
225 - 81=144
12x12=144
15 - 12=3
2006-12-10 13:21:48 · answer #6 · answered by purdygoode 5 · 0⤊ 0⤋
15 squared is 225
12 squared is 144
difference is 81
2006-12-10 13:05:52 · answer #7 · answered by Kerry S 2 · 0⤊ 0⤋
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2016-12-11 06:37:51 · answer #8 · answered by ? 4 · 0⤊ 0⤋
The answer is 93.
2006-12-10 13:07:52 · answer #9 · answered by Justin 1 · 0⤊ 0⤋
9cm2
2006-12-10 13:01:06 · answer #10 · answered by Anonymous · 0⤊ 0⤋