Question

Given: DEFG is a trapezoid DE=FG, DF=a, m∠FDG=45° Find: Area of DEFG

Answer 1

Answer:

Area of DEFG = (1/2)a²

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.

_____

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