The answer is $\boxed{36}$.

Let $a$,$b$,$c$, and $d$ be the side lengths where $a+c=12$ and $b+d=13$. Also, let $m$ and $n$ be the diagonals of the quadrilateral (they are not shown in the diagram below for simplicity). The area of this quadrilateral is $\frac{mn}{2}$. By  Ptolemy's theorem, we have $mn = ac + bd$. Therefore we are trying to find the maximal value of $$\frac{ac+bd}{2}$$

First, let's show that $$a^2+c^2=b^2+d^2$$

For this, we have marked the equal angles $\alpha$ and $\beta$ in the diagram above using concyclic quadrilateral's properties. Now we have $$\left\{ \begin{array}{rcl} a\sin{\alpha} &=& b\cos{\beta}\\ a\cos{\alpha} &=& d\cos{\beta} \\ c\cos{\alpha} &=& d\sin{\beta} \\ c\sin{\alpha} & = &b\sin{\beta} \end{array} \right.$$

Squaring both sides of these four equations and then adding them together gives desired result $a^2 + c^2 = b^2 + d^2$.

Meanwhile, we have $a+c=12$ and $b+d=13$. Therefore, we have $$a^2 + c^2 = b^2+d^2 \implies (a+c)^2-2ac = (b+d)^2-2bd \implies bd - ac = \frac{25}{2}$$

Now the problem becomes equivalent to maximizing the value of $\frac{ac+bd}{2}$ given $$\left\{\begin{array}{rcl} a + c & = & 12 \\ b + d & = & 13 \\ bd - ac & = & \frac{25}{2} \end{array} \right.$$

Because $$(ac+bd)^2 = (bd-ac)^2 +4acbd = \frac{625}{4} + 4acbd$$

therefore $\frac{ac+bd}{2}$ will maximize when the value of $acbd$ maximizes.

By the basic AM-GM inequality, we have  $$ac\le \left(\frac{a+b}{2}\right)^2 = 36 \qquad \text{and} \qquad bd \le \left(\frac{b+d}{2}\right)^2=\frac{169}{4}$$

Let $x = ac$ and $y = bd$. Due to the fact that $a$,$b$,$c$, and $d$ are all positive, we have $0<x\le 36$ and $0<y\le \frac{169}{4}$. Additionally, $y - x = bd-ac=\frac{25}{2}$. The target, $acbd$, will become $xy$.

Replacing $x$ with $\left(y-\frac{25}{2}\right)$, we have $xy=y\left(y-\frac{25}{2}\right)$. The constraint will be $0 < y \le \frac{169}{4}$ and $0 < x= y-\frac{25}{2} \le 36$ which means $0 < y \le \frac{169}{4}$. Hence, we find its maximum can be reached when $y=\frac{169}{4}$. This can also be verified using the geometric method.

In the diagram above, the domain of possible $(x, y)$ are the rectangle bounded by $x= 0$, $x=36$ and $y=0$, $y=\frac{169}{4}$. The constraint means that only those points on the line $y=x+\frac{25}{2}$ should be considered. We are trying to find a hyperbola $xy=k$ which is as far from the origin as possible. Clearly, it must pass the intercept of $y=x+\frac{25}{2}$ and $y=\frac{169}{4}$, which is $(\frac{119}{4},\frac{169}{4})$. Accordingly the maximum area is $$\frac{ac+bd}{2} = \frac{x+y}{2} = \boxed{36}$$

Such a quadrilateral indeed exists. At this point, $a=\frac{7}{2}$, $b=d=\frac{13}{2}$, and $c = \frac{17}{2}$. The four equations at the beginning of the solution about $(a, b, c, d,\alpha,\beta)$ can be solved, which gives $\alpha = \arcsin(\frac{91}{338})\approx 15.62^{\circ}$ and $\beta =\arccos(\frac{49}{338})\approx 81.66^{\circ}$.