In the figure below, $ACEG$ is a rectangle and $ABJH$ is a square. The area of the shaded rectangle $JDEF$ is $36\text{ cm}^2$ and the length of $HG$ is $9\text{ cm}$. Find the length of $AC$.

1. Determine the dimensions of the shaded rectangle $JDEF$.

Since $HD$ is horizontal and $BF$ is vertical, $JF$ is parallel and equal to $HG$. Thus, $JF = 9\text{ cm}$.
The area of $JDEF$ is $36\text{ cm}^2$, so $JD = \frac{36}{9} = 4\text{ cm}$.

2. Relate the lengths to the square $ABJH$ and set up an equation using similar triangles.

Let the side length of the square $ABJH$ be $x\text{ cm}$. Then $AB = AH = HJ = BJ = x\text{ cm}$.
Since $BC = JD = 4\text{ cm}$, the total length $AC = x + 4$.
Since $G, J, C$ are collinear, $\triangle GHJ$ is similar to $\triangle GAC$ (or by comparing gradients).
$\frac{HJ}{AC} = \frac{GH}{GA}$

3. Solve for $x$.

Substitute the known lengths into the similarity ratio:
$\frac{x}{x + 4} = \frac{9}{9 + x}$
Cross-multiplying gives:
$x(9 + x) = 9(x + 4)$
$9x + x^2 = 9x + 36$
$x^2 = 36$
Since $x > 0$, $x = 6$.

4. Calculate the length of $AC$.

$AC = AB + BC = x + 4 = 6 + 4 = 10\text{ cm}$.