Answer:
Area of DEFG = (1/2)a²
Step-by-step explanation:
See the attached diagram.
Line FB ║ EG and line GA ║DF. ∆GFA ≅ ∆FGX, and ∆EXF ≅ ∆BAG. Then the area of ∆BFG is equal to the area of ∆EGF and, by symmetry, the area of ∆FDE.
This means the area of DEFG is the same as the area of ∆DFB.
The problem statement tells us ∠FDG = 45°. By symmetry, ∠EGD = 45° and by corresponding angles related to parallel lines ∠FBD = 45°. Then ∠DFB = 90° and the area of ∆DFB = (1/2)(DF)(FB) = (1/2)a².
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Alternate approach
Recognize that ∠EGD = ∠FDG = 45°, so ∠DXG = 90°. This means all the triangles internal to DEFG are right triangles. Designate the distance XF with the variable b. Then DX = (a-b) and the area of the trapezoid is ...
... Area(DEFG) = Area(DXG) + Area(DXE) + Area(GXF) + Area(FXE)
... = (1/2)(a-b)² +(1/2)(a-b)(b) +(1/2)(a-b)(b) +(1/2)b²
... = (1/2)(a² -2ab +b² +ab -b² +ab -b² +b²)
... = (1/2)(a² +ab(-2+1+1) +b²(1-1-1+1))
... Area(DEFG) = (1/2)a²