There was no change in the rectangle's area [tex](0 $ ft$^2).[/tex]
Let's start out by finding what a [tex]15[/tex][tex]\%[/tex] increase in the length [tex](12 $ ft$)[/tex] is. [tex]25\%,[/tex] or [tex] \frac{1}{4} [/tex] of [tex]12[/tex] is [tex]3[/tex]. Since the length was increased by that much, you add the [tex]3[/tex] to [tex]12[/tex] to get a longer length of [tex]15 $ ft$.[/tex]
The same goes for the width: [tex]20\%,[/tex] or [tex] \frac{1}{5}[/tex] of [tex]5[/tex] is [tex]1.[/tex] However, since the width was decreased, we'll subtract this value from [tex]5[/tex], for a shorter width of [tex]4 $ ft$.[/tex]
We've found that the new rectangle is [tex]15 $ ft$[/tex] long and [tex]4 $ ft$[/tex] wide. Its area, which is its length [tex]\times[/tex] width, is [tex]15 \times 4,[/tex] or [tex]60 $ ft$^2.[/tex]
To find how much the area changed, simply subtract this value from the original area of the rectangle [tex](12 \times 5 = 60 $ ft$^2).[/tex]
[tex]60 $ ft$^2 - 60 $ ft$^2 = 0 $ ft$^2[/tex]
We've now found that there was (surprisingly) no change in the rectangle's area, despite changing its size.
