Question 827082: The perimeter of a square is 20 ft. If you increase the length of the square by 2 feet and decrease the width by 1 foot, what is the area, in square feet, of the new figure?
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website! x and y for original square.
2x+2y=20 and A=xy, A for area.
x+y=10 so solving for y, y=10-x.
A=x(10-x).
Note, this is a square, so x=y.
A=x^2;
4x=20 ------ perimeter for square is x+x+x+x=4x
 , the original value for x, for the square.
Changed area, now rectangle.
x+2 and x-1 for this rectangle.
Let B = this new area of this rectangle.
B=(x+2)(x-1)
B=x^2+x-2
We already found that x=5, so just compute B using x=5.
B=5*5+5-2
B=25+5-2
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