6. Let $A B C D$ be a convex quadrilateral, $A B=7, B C=4$, $C D=5, D A=6$, and its area $S$ has a range of $(a, b]$. Then $a+b=$ $\qquad$

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$6.2 \sqrt{210}$.
As shown in Figure 4, connect $B D$, and let
$\angle B A D=a, \angle B C D=\beta$, then
$S=21 \sin a+10 \sin \beta$.
By the cosine rule, we have
$$
\begin{array}{l}
6^{2}+7^{2}-2 \times 6 \times 7 \cos a \\
\quad=B D^{2}=4^{2}+5^{2}-2 \times 4 \times 5 \cos \beta,
\end{array}
$$

which means $21 \cos a-10 \cos \beta=11$.
$(1)^{2}+(2)^{2}$, we get
$$
S^{2}+121=21^{2}+10^{2}+420(\sin \alpha \cdot \sin \beta-\cos \alpha \cdot \cos \beta).
$$

Therefore, $S^{2}=420-420 \cos (\alpha+\beta) \leqslant 840$, which means $S \leqslant 2 \sqrt{210}$, with equality holding if and only if $\alpha+\beta=180^{\circ}$.

Since $A D+C D=A B+B C=11$, when $A C \rightarrow 11$, $S \rightarrow 0$.

In summary, the range of the area $S$ is $(0,2 \sqrt{210}]$, hence $a+b=2 \sqrt{210}$.