The point of intersection of the diagonals of a rectangle is 4 cm further away from the shorter side than from the longer side of the rectangle. The perimeter of the rectangle is equal to 56 cm. Find the lengths of the sides of the rectangle.
Let the shorter side of the rectangle have length $x$, and let the longer side have length $y$. Thus, the distance from the point of intersection of the diagonals to the shorter side is $(x+4)/2 = (x/2) + 2$ cm, and the distance from the point of intersection of the diagonals to the longer side is $(y+4)/2 = (y/2) + 2$ cm.
Since opposite sides of a rectangle have the same length, we have $2x + 2(y/2 + 2) = 56$, or $2x + y + 4 = 56$. This reduces to $2x+y = 52$.
Since the diagonals of a rectangle are of equal length, the point of intersection of the diagonals is equidistant from the four vertices of the rectangle. Thus, the point of intersection is midpoint of the segment connecting the midpoint of each side, giving $x = y/2$ by symmetry.
Substituting $x = y/2$ into the equation $2x+y = 52$ gives $2(y/2) + y = 52$, or $2y + y = 52$, which yields $y = 26$. Hence, $x = (y/2) = (26/2) = 13$, so the dimensions of the rectangle are $\boxed{13 \text{ cm} \times 26 \text{ cm}}$.