The area of a square is equal to the area of a rectangle, which has one of its sides equal to thrice the length of a side of the square.

Let's say that x = length of the sides of the square, y = side of the rectangle that isn't 3x.

Since we're given that the areas of the square and rectangle are equal, we can set up this equation:
\(x*x = (3x)(y)\)
\(y = \frac{x*x}{3x} = \frac{1}{3}x\)

Then we plug in for columns A and B:
Twice the perimeter of the square \(= 8x\)
Perimeter of the rectangle \(= 2(3x+y) = 2(3x+\frac{1}{3}x) = \frac{20}{3}x\)

Since \(8 > \frac{20}{3}\), twice the perimeter of the square must be greater than perimeter of the rectangle.