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