Question 1199408: A square poster is replaced by a rectangular poster that is 2 inches wider and 2 inches shorter. What is the difference in the number of square inches between the area of the larger poster and the smaller poster?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
x = side length of the original square
This is some positive real number.
The rectangular poster is 2 inches wider.
This means x becomes x+2 along the width.
Meanwhile, the height goes from x to x-2 because it's 2 inches shorter.
The square poster is x inches by x inches.
The rectangular poster is (x+2) inches by (x-2) inches.
A = area of the square = side*side = x*x = x^2
B = area of the rectangle = width*height = (x+2)(x-2) = x^2-4
Use the difference of squares rule to see why (x+2)(x-2) = x^2-4.
In short
A = x^2
B = x^2-4
Subtract these items to find that
A - B = (x^2) - (x^2-4) = x^2-x^2+4 = 4
So
A-B = 4
The difference in area is 4 square inches.
The square has the larger area because we subtracted off 4 in the area of the rectangle.
Put another way:
A = area of the square = x^2
B = area of the rectangle = x^2-4 = A-4
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If you wanted, you could do a numeric example to help cement things better.
Let's say the square poster is 10 inches by 10 inches.
Feel free to pick whatever positive number you prefer.
The area is 10*10 = 100 square inches.
The rectangular poster would be 10+2 = 12 inches by 10-2 = 8 inches
The area of this is 12*8 = 96 square inches.
Therefore, the difference in areas is 100-96 = 4 square inches.
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Answer: 4 square inches
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