Given: defg is a trapezoid de=fg, df=a, m∠fdg=45° find: area of defg

Area of DEFG = (1/2)a²

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

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

pentagon

step-by-step explanation:

the coordinates   a(5, -2), b(3,-1),c(-4,-4) d(-3,8) and e(-1,4) forms the vertices of the polygon when they are connected in order.

in order to classify that which polygon is it with the given coordinates, first count the number of vertices and then count the number of sides it can make with these vertices.

since, the given coordinates have five vertices, therefore, it will make five sides. moreover, pentagon is regular in nature therefore it will be convex.

now, a polygon with five sides is known as pentagon, therefore, the given polygon is the pentagon.