Answer:
12 unit²
Explanation:
area ΔAMC = 1/2 CM x AH' = 1/4 BC x AH'
area ΔABM = 1/2 BM x AH' = 1/4 BC x AH'
area ΔAMC = area ΔABM
for the same calculation, we can prove
area ΔACN = area ΔDCN
area ΔAMC + area ΔACN = 6
area ΔABM + area ΔDCN = 6
ABCD = area ΔAMC + area ΔACN + area ΔABM + area ΔDCN = 6 = 6 = 12