Question 827082
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
{{{highlight(x=5)}}}, 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
{{{highlight(B=28)}}}