Let a equal the area of the original rectangle, l equal length and w equal width.
The original rectangle would have been [tex]l*w= a[/tex]
The modified rectangle would have been [tex](1.1l)*(1- \frac{p}{100})w=.88a
This can be rewritten as lw(1.1)(1-p/100)=.88a
Since a= lw, we can divide both sides of the function a if we substitute
So we get (1.1)(1-p/100)=.88
(1-p/100)=.8
100-p=80
p=20
[/tex]
So the value of p is 20
