Question 1138434: Each side of a square is increased 9 inches. When this happens, the area is multiplied by 16. How many inches in the side of the original square?
Found 3 solutions by ikleyn, Theo, greenestamps:
Answer by ikleyn(52778)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! each side of the square is increased by 9 inches.
the area is multiplied by 16.
let x = the side of the square.
the area of the square is therefore x^2.
when the side is increased by 9 inches, the area becomes (x+9)^2 = x^2 + 18x + 81.
you get x^2 + 18x + 81 = 16x^2
x^2 - 18x + 81 is the area when the side is increased by 9 inches.
16x^2 is the size of the new area which is 16 times the size of the original area.
subtract the left side of this equation from both sides of this equation to get:
16x^2 - x^2 = 18x - 81 = 0
combine like terms to get 15x^2 - 18x - 81 = 0
solve this quadratic equation by using the quadratic formula to get x = -1.8 or x = 3.
x has to be positive, so the solution has to be 3 or not at all.
when x = 3, x^2 = 9
when x = 3 + 9 = 12, x^2 = 144.
144 = 16 * 9, confirming the solution is correct.
your solution is that the original length of the size is 3 inches.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
All squares are similar.
If the ratio of the areas of two squares is 1:16, then the ratio of the lengths of the sides of the two squares is 1:4.
If increasing the side length by 9 inches multiplies the side length by 4, then those 9 inches are 3 times the original side length; so the original side length is 9/3 = 3 inches.
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