2006 AMC 10A Problem 7

Start with what cannot change when you cut and move pieces: the total area.

1. Size of the square. The rectangle has area $8 \times 18 = 144$. The square is made of the same two pieces, so its area is also $144$, and its side is $12$ (because $12 \times 12 = 144$).
2. Read the figure. The cut starts on the top edge, goes down $4$ (half the height of $8$), goes across, then goes down another $4$ to the bottom edge. The left hexagon has a short top edge of length $y$ and a long bottom edge of length $18 - y$. The right hexagon is the same shape turned upside down.
3. Build the square. Lift the right hexagon up by $4$ and slide it left so the two "steps" of the cut fit together. The height becomes $8 + 4 = 12$, which matches the square. The long edges of the hexagons are now the top and bottom sides of the square, so
$$18 - y = 12, \qquad y = 6.$$
4. Check. The middle, horizontal part of the cut has length $18 - y - y = 18 - 12 = 6$, which equals $y$. So the short edges fit exactly into the notches, with no gaps.

The answer is $\boxed{\textbf{(A)}\ 6}$.

In any cut-and-rearrange problem, area stays the same, and that usually tells you the size of the new shape right away. Then match one edge of a piece to a side of the new shape to get an equation. Checking a second edge confirms that the pieces really fit.

For the two steps to lock together, the three horizontal pieces along the length $18$ (the top edge $y$, the middle of the cut, and the bottom edge $y$ of the other piece) must all be the same length. Three equal pieces make $18$, so $y = 18 \div 3 = 6$.