Circles with centers at $O$ and $P$ have radii 2 and 4, respectively, and are externally tangent. Points $A$ and $B$ on the circle with center $O$ and points  $C$ and $D$ on the circle with center $P$ are such that  $\overline{AD}$ and $\overline{BC}$ are common external tangents to the circles. What is the area of the concave hexagon $AOBCPD$?

[asy]
pair A,B,C,D,H,O,P;
O = (0,0);
P = (6,0);
H = (-6,0);
A = intersectionpoint(arc((A + H)/2, abs(A -  H)/2, 0, 180), arc(O, 2, 0, 180));
D = intersectionpoint(arc((A + P)/2, abs(A -  P)/2, 0, 180), arc(P, 4, 0, 180));
B = reflect(O,P)*(A);
C = reflect(O,P)*(D);
draw(O--P,linewidth(0.7));
draw(Circle(O,2),linewidth(0.7));
draw(Circle(P,4),linewidth(0.7));
draw(interp(A,D,-0.2)--interp(A,D,1.2),linewidth(0.7));
draw(interp(B,C,-0.2)--interp(B,C,1.2),linewidth(0.7));
draw(A--O--B,linewidth(0.7));
draw(D--P--C,linewidth(0.7));
label("$O$",O,SE);
label("$P$",P,E);
label("$D$",D,NW);
label("$A$",A,NW);
label("$B$",B,S);
label("$C$",C,S);
label("2",(-0.45,0.95),E);
label("6",(3,0),N);
label("4",(5.55,1.95),E);

[/asy]
Through $O$ draw a line parallel to $\overline{AD}$ intersecting $\overline{PD}$ at $F$. [asy]
pair A,B,C,D,H,O,P,F;
O = (0,0);
P = (6,0);
H = (-6,0);
A = intersectionpoint(arc((A + H)/2, abs(A -  H)/2, 0, 180), arc(O, 2, 0, 180));
D = intersectionpoint(arc((A + P)/2, abs(A -  P)/2, 0, 180), arc(P, 4, 0, 180));
B = reflect(O,P)*(A);
C = reflect(O,P)*(D);
F=(5.5,1.95);
draw(O--P,linewidth(0.7));
draw(Circle(O,2),linewidth(0.7));
draw(Circle(P,4),linewidth(0.7));
draw(interp(A,D,-0.2)--interp(A,D,1.2),linewidth(0.7));
draw(interp(B,C,-0.2)--interp(B,C,1.2),linewidth(0.7));
draw(A--O--B,linewidth(0.7));
draw(D--P--C,linewidth(0.7));
label("$O$",O,SE);
label("$P$",P,E);
label("$D$",D,NW);
label("$A$",A,NW);
label("$B$",B,S);
label("$C$",C,S);
label("2",(-0.45,0.95),E);
label("6",(3,0),N);
label("$F$",(5.55,1.95),E);
draw(O--F,linewidth(0.7));
label("2",(5.3,2.9),W);
label("2",(5.7,1),W);

[/asy] Then $AOFD$ is a rectangle and $OPF$ is a right triangle. Thus $DF=2$, $FP=2$, and $OF=4\sqrt{2}$. The area of trapezoid $AOPD$ is $12\sqrt{2}$, and the area of hexagon $AOBCPD$ is $2\cdot 12\sqrt{2}=\boxed{24\sqrt{2}}$.