Problem: The side length of the largest square below is $8\sqrt{2}$, as shown. Find the area of the shaded region.

\[ \text{[asy]} \]
\text{size}(10\text{cm}); \text{real lsf}=0.5; \text{pen dps}=\text{linewidth}(0.7)+\text{fontsize}(10); \text{defaultpen}(dps); \text{pen ds}=\text{black}; \text{real xmin}=-18.99425911800572, \text{xmax}=23.81538435842469, \text{ymin}=-15.51769962526155, \text{ymax}=6.464951807764648; 
\text{pen zzttqq}=\text{rgb}(0.6,0.2,0.); 
\text{pair A}=(0.,1.), \text{B}=(0.,0.), \text{C}=(1.,0.), \text{D}=(1.,1.), \text{F}=(1.,2.), \text{G}=(2.,3.), \text{H}=(0.,3.), \text{I}=(0.,5.), \text{J}=(-2.,3.), \text{K}=(-4.,5.), \text{L}=(-4.,1.), \text{M}=(-8.,1.), \text{O}=(-8.,-7.), \text{P}=(0.,-7.); 
\text{draw}(B--A--D--C--\text{cycle}); \text{draw}(A--C--(2.,1.)--F--\text{cycle}); \text{draw}(A--(2.,1.)--G--H--\text{cycle}); \text{draw}(A--G--I--J--\text{cycle}); \text{draw}(A--I--K--L--\text{cycle}); \text{draw}(A--K--M--(-4.,-3.)--\text{cycle}); \text{draw}(A--M--O--P--\text{cycle}); \text{draw}(A--O--(0.,-15.)--(8.,-7.)--\text{cycle}); \text{filldraw}(A--B--C--D--\text{cycle},\text{opacity}(0.2)+\text{black}); \text{filldraw}(A--(2.,1.)--F--\text{cycle},\text{opacity}(0.2)+\text{black}); \text{filldraw}(A--G--H--\text{cycle},\text{opacity}(0.2)+\text{black}); \text{filldraw}(A--I--J--\text{cycle},\text{opacity}(0.2)+\text{black}); \text{filldraw}(A--K--L--\text{cycle},\text{opacity}(0.2)+\text{black}); \text{filldraw}(A--M--(-4.,-3.)--\text{cycle}, \text{opacity}(0.2)+\text{black}); \text{filldraw}(A--O--P--\text{cycle},\text{opacity}(0.2)+\text{black}); 
\text{draw}(B--A); \text{draw}(A--D); \text{draw}(D--C); \text{draw}(C--B); \text{draw}(A--C); \text{draw}(C--(2.,1.)); \text{draw}((2.,1.)--F); \text{draw}(F--A); \text{draw}(A--(2.,1.)); \text{draw}((2.,1.)--G); \text{draw}(G--H); \text{draw}(H--A); \text{draw}(A--G); \text{draw}(G--I); \text{draw}(I--J); \text{draw}(J--A); \text{draw}(A--I); \text{draw}(I--K); \text{draw}(K--L); \text{draw}(L--A); \text{draw}(A--K); \text{draw}(K--M); \text{draw}(M--(-4.,-3.)); \text{draw}((-4.,-3.)--A); \text{draw}(A--M); \text{draw}(M--O); \text{draw}(O--P); \text{draw}(P--A); \text{draw}(A--O); \text{draw}(O--(0.,-15.)); \text{draw}((0.,-15.)--(8.,-7.)); \text{draw}((8.,-7.)--A); \text{draw}(A--B,\text{black}); \text{draw}(B--C,\text{black}); \text{draw}(C--D,\text{black}); \text{draw}(D--A,\text{black}); \text{draw}(A--(2.,1.),\text{black}); \text{draw}((2.,1.)--F,\text{black}); \text{draw}(F--A,\text{black}); \text{draw}(A--G,\text{black}); \text{draw}(G--H,\text{black}); \text{draw}(H--A,\text{black}); \text{draw}(A--I,\text{black}); \text{draw}(I--J,\text{black}); \text{draw}(J--A,\text{black}); \text{draw}(A--K,\text{black}); \text{draw}(K--L,\text{black}); \text{draw}(L--A,\text{black}); \text{draw}(A--M,\text{black}); \text{draw}(M--(-4.,-3.),\text{black}); \text{draw}((-4.,-3.)--A,\text{black}); \text{draw}(A--O,\text{black}); \text{draw}(O--P,\text{black}); \text{draw}(P--A,\text{black}); \text{label}("$8\sqrt{2}$",(-8,-7)--(0,-15));
\text{clip}((\text{xmin},\text{ymin})--(\text{xmin},\text{ymax})--(\text{xmax},\text{ymax})--(\text{xmax},\text{ymin})--\text{cycle});
Answer: 64