SOLUTION: I need help solving please:
When the length of a square is increased by 6 and the width is decreased by 4, the area remains unchanged. Find the dimensions of the square.
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When the length of a square is increased by 6 and the width is decreased by 4, the area remains unchanged. Find the dimensions of the square.
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Question 249008: I need help solving please:
When the length of a square is increased by 6 and the width is decreased by 4, the area remains unchanged. Find the dimensions of the square.
Thank you!! Found 2 solutions by dabanfield, Earlsdon:Answer by dabanfield(803) (Show Source):
You can put this solution on YOUR website! When the length of a square is increased by 6 and the width is decreased by 4, the area remains unchanged. Find the dimensions of the square.
Let x be the length of a side of the square.
Then x*x = x^2 is the area of the square.
From the givens we have:
(x+6)*(x-4) = x^2
Expanding we have:
x^2 + 2*x - 24 = x^2
Simplifying:
2*x = 24
x = 12
You can put this solution on YOUR website! Let the length of the side of the original square be x.
The area of the original square is:
The area of the new rectangle is: But this is equal to the area of the original square, so... Subtract from both sides. Add 24 to both sides. Divide both sides by 2.
The dimensions of the square are: 12 X 12
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