Having the side lengths alternate $6$, $4$, $6$, $4$,$\dots$, we can split this octagon into $4$ triangles, $1$ square, and $4$ rectangle. Each triangle has an equal area, and because they are right with a hypotenuse of $6$, the side lengths of each of them are $3\sqrt2$. Thus, the area all together is $4\cdot3\sqrt2\cdot3\sqrt2\cdot\frac12=36$ The rectangles have a height of $3\sqrt2$ and a width of $4$, so the area of all four of them is $4\cdot4\cdot3\sqrt{2}=48\sqrt2$. The square in the middle has an area of $4^2=16$, and so altogether the area of the octagon is $36+16+48\sqrt2=52+48\sqrt2.$

Given a cyclic octagon with any sidelenghts, it doesn't matter the order of the sides, by imagining the octagon as slices of pizza (circular sectors) of octagon. If it fits in a circle, by reordering the slices it should also fit, but in the same circle.

Split the octagon into 8 triangles, each has 2 vertices on the octagon and one at the center. Consider any pair of $6$ and $4$ triangle. If $r$ is the circumradius, then $$arg((2+\sqrt{r^2-4}i)(3+\sqrt{r^2-9}i))=\frac{3\pi}{4}$$$$Re((2+\sqrt{r^2-4}i)(3+\sqrt{r^2-9}i))=-Im((2+\sqrt{r^2-4}i)(3+\sqrt{r^2-3}i))$$$$6-\sqrt{(r^2-4)(r^2-9)}=-3\sqrt{r^2-4}-2\sqrt{r^2-9}$$If $a=\sqrt{r^2-9}$, then we have $\sqrt{r^2-4}=\sqrt{a^2+5}$, $$6-a\sqrt{a^2+5}=-3\sqrt{a^2+5}-2a$$$$6+2a=(a-3)\sqrt{a^2+5}$$$$4a^2+24a+36=(a^2-6a+9)(a^2+5)$$$$4a^2+24a+36=a^4-6a^3+14a^2-30a+45$$$$a^4-6a^3+10a^2-54a+9=0$$Using MUD and assuming that it factors into the product of 2 quadratic terms, $$(a^2-6a+1)(a^2+9)=0$$Since $a>0$, $$a=3\pm 2\sqrt{2}$$Since $\sqrt{a^2+5}=\frac{6+2a}{a-3}\geq 0$, we require that $a>3$, which means our only solution is $$a=3+2\sqrt{2}$$The altitudes to the $6$ side and the $4$ side are respectively, $$\sqrt{r^2-9}=3+2\sqrt{2}$$$$\sqrt{r^2-4}=\sqrt{22+12\sqrt{2}}=2+3\sqrt{2}$$The area of the polygon is $$4\cdot\frac{1}{2}(4\sqrt{r^2-4}+6\sqrt{r^2-9})$$$$2(8+12\sqrt{2}+18+12\sqrt{2})$$$$52+48\sqrt{2}$$