12. As shown in Figure 4, in the octagon $A B C D E F G H$, the areas of 4 squares are $25$, $144$, $48$, and $121$ square units, and $P R=13$ (units). Then the area of the octagon $=$ $\qquad$ square units.
Obviously, $S_{\triangle M B C}+S_{\triangle C D Q}$
As shown in Figure 10, draw $Q I \perp P S$ intersecting $P S$ at point $I$, and $B J \perp A P$ intersecting the extension of $A P$ at point $J$. It is easy to prove that Rt $\triangle P Q I \cong \mathrm{Rt} \triangle P B J$, thus $Q J = B J$. Therefore, $S_{\triangle A P B}=S_{\triangle \text { SQ }}$ (equal base and height).
Similarly, $S_{\triangle E F K}=S_{\text {DNSK }}$.
Then, $S_{\triangle P P B}+S_{\triangle E F R}=S_{\text {quadrilateral } \operatorname{LNNT}}$.
Thus, the area of the octagon
$$
=3(30+22 \sqrt{3})+144+48+12+25=428+66 \sqrt{3} \text {. }
$$