The $8\times18$ [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle) $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square.  What is $y$?

$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$
Since the two [hexagons](https://artofproblemsolving.com/wiki/index.php/Hexagon) are going to be repositioned to form a [ square](https://artofproblemsolving.com/wiki/index.php/Square_(geometry)) without overlap, the [area](https://artofproblemsolving.com/wiki/index.php/Area) will remain the same. The rectangle's area is $18\cdot8=144$. This means the square will have four sides of length 12. The only way to do this is shown below.

As you can see from the diagram, the [line segment](https://artofproblemsolving.com/wiki/index.php/Line_segment) denoted as $y$ is half the length of the side of the square, which leads to $y = \frac{12}{2} = \boxed{\textbf{(A) }6}$.

Solution 1  (Shortcut)
As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.


As you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$.


From there we can say $3y = 18$ so $y = \frac{18}{3} = \boxed{\textbf{(A) }6}$.
~Ezraft