Shadow area = area of triangle BCH-area of small rectangle CEFH
=6×4÷2-4×(6-4)= 12-8=4
The overlapping part (common part) of two triangles is a small triangle CDE, so the two triangles are equal.
So the shadow area in the figure = trapezoidal ABCD area = [(10-3)+10] × 2 ÷ 2 =17.
Shadow area = half of small rectangle area = quarter of whole (large) rectangle area.
9×8÷4= 18
4×(6+df)÷2-[4+﹙4-be﹚]×6÷2-6×be÷2=6
12+2DF-24+3BE-3BE=6
2DF- 12=6
2DE= 18
DF=9
The answer can be obtained through translation.
24×2+ 16×2-2×2 =76
Shadow area = triangular BCD area-triangular BEF area
=24×2/3-24÷2÷2
= 16-6
= 10