please help with algebra question?

Question

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

Answer 1

Let x cm be the original dimension
original area = x^2
new area = (x - 3)(x - 3) = x^2 -6x + 9
new area = original area - 81
x^2 - 6x + 9 = x^2 - 81
=> x = 15 cm
=> the original was a 15 cm by 15 cm square

proof:
original area = 15^2 = 225 cm^2
new area = (15 - 3)^2 = 12^2 = 144 cm^2
original area - new area = 225 - 144 = 81 cm^2

Answer 2

Let s be the side of the original square. Then its area is
A = s x s = s^2
Now, the side of the new square is s' = s-3, then its area is
A' = s' ^2 = (s-3)^2 = s^2 - 6s + 9
The difference between both area is
A - A' = s^2 - (s^2 - 6s + 9) = 81
s^2 - s^2 + 6s - 9 = 81
6s - 9 = 81
6s = 90
s = 15
Check:
A = 15 ^ 2 = 225
A' = 12 ^ 2 = 144
225 - 144 = 81 correct.

Answer 3

Answer is 15 cm by 15 cm.

If side of a square is x, then x^2-(x-3)^2=81.

On solving, x^2-x^2-9+6x=81,

means x=15 cm

Answer 4

Let x = the orignal side of the square
Let A = the original area of the square
A = x^2

Decreased Square:
(x-3)(x-3) = A - 81 = x^2 - 81
x^2 - 6x + 9 = x^2 - 81
-6x + 9 = -81
-6x = -90
x = 15

Therefore:
The original dimensions of the square were 15cm by 15cm

Answer 5

Area of square =( L * L) where L is length of side
Area of small square is (L-3) * (L-3)

Difference in areas is 81 = (L*L) minus (L-3)(L-3)
81 = L2 - (L2-6L+9) (read L2 as L squared)
81 = 6L - 9
therefor 90 = 6L
so Length =90/6 =15 cm

Answer 6

The dimensions of the original square were 15 cm by 15 cm

Let x = a side of the original square
Then x*x - (x-3)(x-3) = 81
x*x - x*x + 6x - 9 = 81
6x = 90
x=15

Answer 7

suppose measure of a side of the original square was x.
then area of square is x^2.if we decrease a side of square by 3
the area of square is decreased by 81 cm2.
so,the equation will be (x-3)^2=x^2-81.solving this equation gives x=15.So,the measure of a side of the original square was 15 and its area was 225cm^2.you can check it.

Answer 8

By decreasing the sides of the square, we decrease its area by one little square (3" x 3") plus two rectangles. Draw it and see.

If x is the side of the original square, the area of each rectangle is 3 * (x - 3). The area of the two rectangles is 6x - 18.

So

6x - 18 + 9 = 81
6x - 9 = 81
6x = 90
x = 15

Answer 9

Let the side of the original square be x.

So the equation you form is x^2 - 81 = (x-3)^2

Expanding, x^2 - 81 = x^2 - 6x + 9
Rearranging, 6x = 90
So x = 15

The dimensions of the original square are 15cm by 15cm

Answer 10

Let the original sides be X cm, with an area of X cm squared.
Then the new sides are X-3 cm, with an area of (X-3)(X-3) sqd.
Therefore, X sqd -(X-3)(X-3) =81
X sqd - (X sqd-6X+9) = 81
Xsqd - X sqd +6X - 9 =81
6X -9 = 81
6X =81 + 9 = 90
X = 90/6 =15 cm