解答:解:设AB=a,FP=b,延长PK,BE交于点M,
∵四边形ABCD为正方形, ∴AB=AD=CD=BC=a, ∴S△AED=(4+a)a, ∵CG=BC-BG=a-4, ∴S△CGD=(a-4)a, ∵四边形FPRK为正方形, ∴FR=RK=PK=FP=b, ∵GF=4, ∴S△KPG=(4+b)b, ∵四边形FEBG、FPKR为正方形, ∴∠MBG=∠BGP=∠P=90°, ∴矩形FPME, ∴PM="4" KM=4-b, ∵EM=b, ∴S△EKM=(4-b)b, ∴S△DKE=(S正方形ABCD+S正方形GFEB+S矩形FPME)-(S△AED+S△CGD+S△GPK+S△EMK), =(a2+42+4b)-[(4+a)a+(a-4)a+(4+b)b+(4-b)b], =a2+16+4b-[2a+a2+a2-2a+2b+b2+2b-b2] =a2+16+4b-[a2+4b] =16; |