Solution #12

Note that triangle ABD and triangle AA'B have bases of the same length (since AD = AA' = 6). Their altitudes from B are also the same length, so they have the same area.

Because AB = BB', the altitude from B' to AD is twice the length of the altitude from B to AD. So triangle AA'B' as twice the area of triangle ABD.

By similar reasoning:

The area of triangle BB'C is twice the area of triangle BAC.
The area of triangle CC'D is twice the area of triangle CBD.
The area of triangle DD'A is twice the area of triangle DCA.

The area of A'B'C'D' is: [AA'B] + [BB'C] + [CC'D] + [DD'A] + [ABCD]

From the above, this is:

2[ABD] + 2[BAC] + 2[CBD] + 2[DCA] + 10

But [ABD] + [CBD] = [ABCD] = 10, and [BAC] + [DCA] = [ABCD] = 10. So:

[A'B'C'D'] = 2*10 + 2*10 + 10 = 50.