I've seen it before - 6

Let Z be the center of ABCD. There are symmetrys about two diagonals and two midpoint lines.
So DHZG is also a 9X9 square, $\frac1 4$ of ABCD, with P and J midpoints of the respective sides.
$\therefore \color{#3D99F6}{HP=PZ=\frac 9 2.\\ \text{With X on GP draw JX || PZ,........... with Y on GZ draw IY || PZ, }\\ \text{with Q on PZ draw XQ}~ \bot ~ PZ, ~~~~~~~R=XQ ~\cap ~ IY.} ~ all ~ on ~sketch.\\ \Delta ~ GPZ, ~J ~is~midpoint ~ of ~ GZ, ~~\therefore ~ ~\color{#3D99F6}{XJ=\dfrac {PZ} 2 =\frac 9 4.}\\ HZ \bot ~XQ ~ and ~ GZ ~~\therefore ~ XQ || JZ ~~and ~~ XJ || QZ || RY ~~~\therefore \color{#3D99F6}{RY=XJ= \frac 9 4.}\\ Since ~XP \cap JH ~~ and ~~XJ||HP, ~~\Delta ~ IJX \text{~} \Delta ~ IHP, ~~\therefore \dfrac{XI}{IP}=\dfrac{XJ}{HP}=\frac 1 2.\\$ $But ~ in ~\Delta ~ XPQ, ~ IR||PQ ~ \implies ~\dfrac{IR}{PQ}=\dfrac a {a+2a}, ~ \therefore ~\color{#3D99F6}{IR=\dfrac{\frac 9 4}3=\frac 3 4.} \\ \therefore ~height ~of ~\Delta ~IZJ ~\color{#3D99F6}{IY=IR+RY=3.}\\ \therefore ~ Area ~ of ~ \Delta ~JIZ ~ = \frac 1 2 +JZ*IY=\dfrac {27}4.\\ But~ \because ~ of ~ symmetry, ~Area ~ of ~octagon ~= 8~ times~ \Delta ~JIZ.\\ Area ~ of ~octagon =\Large ~~~~~\color{#D61F06}{54}$
In spite of symmetry, adjacent angles of the octagon are not equal, only alternative angles are equal.

Uses coordinate geometry.

Set $A$ to be the origin.

We can prove that the octagon is regular (will leave it to the reader)

Also, we know that the center of the octagon is the center of the square. (Again, up to the reader to prove) Thus, its center is $(9,9)$ .

Take any vertex of the octagon. So let's say it's the bottom one (Point $M$ I think). The coordinates of the point is simply the intersection between the line that passes through A and the midpoint of $CD$ . Knowing that the side length of the square is 18, we then know that $D=(18,0)$ . Solving this, we have the equations $y=\frac{1}{2}x$ and $y=-\frac{1}{2}x+9$ . Therefore their intersection point, and one of the octagon's vertices, is $\left(9, \dfrac{9}{2}\right)$ Doing the same thing to any the next vertex (counter-clockwise), we have it's coordinates being $(12,6)$ . We can now compute for the triangle's area. Which is conveniently $\frac{1}{8}$ of the octagon's area.

We can now calculate the triangle's area. Its area is then $\frac{1}{2} \begin{vmatrix} 9 & 9 \\ 9 & \frac{9}{2} \\ 12 & 6 \\ 9 & 9 \end{vmatrix}=\frac{1}{2}\mid\left(\frac{81}{2}+51+108\right)-(81+54+54)\mid=\frac{27}{4}$ .

Thus, the area of the rectangle is $\frac{27}{4}\times 8=\boxed{54}$ .

I got a little different solution.

Also, let $A=(0,0)$ . Since the octagon is regular, it's sufficent to find one side. Let's find the side $MN$ . To do so, we'll find the equation of the lines $AF$ and $BH$ . They are $y=\dfrac{x}{2}$ and $y=-\dfrac{x}{2}+9$ , respectively. And their intersection point is $N=\left(9,\dfrac{9}{2}\right)$ .

Now, do the same with the lines $AF$ and $CE$ . They are $y=\dfrac{x}{2}$ and $y=2x-18$ , respectively. Their intersection point is $M=(12,6)$ .

Finally, calculate the side $MN$ with the distance formula: $MN=\sqrt{(12-9)^2+\left(6-\dfrac{9}{2}\right)^2}=\sqrt{9+\dfrac{9}{4}}=\dfrac{3\sqrt{5}}{2}$ .

Use the formula of the area of the octagon:

$Area=2(\sqrt{2}+1)L^2=2(\sqrt{2}+1)\dfrac{45}{4}=\dfrac{45+45\sqrt{2}}{2}\approx 54.31$ , which is very near from 54. Is my approach correct?