Let the original side of the square be \(s\), so its original area is \(s^2\). When the side is decreased by 50%, the new side becomes \(0.5s\), and the new area is \((0.5s)^2 = 0.25s^2\), just 25% of the original.
That means the area has fallen by \(100\% - 25\% = 75\%\).
Since area depends on the square of the side length, it drops much faster than the side itself, which is why halving the side causes such a large 75% decrease. That matches option 2.