In the second picture, the height ratio of the parallelogram EFGH to ABHG is 5:6 (cross E is the vertical line of AB, which can be easily proved by similar triangles), so the area ratio of the parallelogram EFGH to ABHG is 5:6 (the same bottom).
The area of triangle ABH is equal to half of the area of parallelogram ABHG, so the area of triangle ABH = 165438+ 0/3 of the area of parallelogram ABEF. ?
The area of parallelogram ABCD is equal to the area of parallelogram ABEF (same base and same height).
Therefore, shadow area = parallelogram area ABEF- triangle area ABH = parallelogram area abefx8/1.
But the area of parallelogram ABEF cannot be determined. The angle is uncertain and the height cannot be determined.