12. The locations of square ABCD, square BEFG and square RKPF are shown in Figure 4. G point is on DK line, and the side length of square BEFG is 4, so the area of triangle DEK is:

10(B) 12(C) 14(D) 16

Solution (d)

Let AB=X, FP=Y, extend PK, at point m.

∵ quadrilateral ABCD is a square.

∴AB=AD=CD=BC=X

∴S△AED=[(4+X)X]/2

∫CG = BC-BG = X-4

∴S△CGD=[(X-4)X]/2

∵ Quadrilateral FPRK is a square.

∴FR=RK=PK=FP=Y

GF = 4

∴S△KPG=[(4+Y)Y]/2

∫ quadrilateral FEBG and FPKR are squares.

∴∠MBG=∠BGP=∠P=90

∴ rectangular FPME

∴PM=4 KM=4-Y

EM = Y

∴S△EKM=[(4-Y)Y]/2

∴S△DKE=(S square ABCD+S square GFEB+S rectangular FPME)-(S△AED+

S△CGD+S△GPK+S△EMK)=(X∧2+ 16+4Y)-{[(4+X)X]/2+[(X-4)X]/2+[(4+Y)Y]/2+[(4-Y)Y]/2 }

Simplified S△DKE= 16