Question

已知,在正方形ABCD中,AB=4,点E在边BC上,BE=3,DF⊥AE,F是垂足,求四边形CDEF的面积

Answer 1

∵四边形ABCD为正方形,DF⊥AE

∴∠ADF+∠FAD=∠FAD+∠BAE=90°,∠DFA=∠ABE=90°

∴∠ADF=∠BAE

∴△ADF∽△EAB

∴AF/EB=AD/AE=DF/AB

又∵AB=4,BE=3

∴AE=5

∴AF/3=4/5=DF/4

∴AF=12/5,DF=16/5

又∵S四边形CDFE=S正ABCD-S△AED-S△ABE

∴S四边形CDFE=4*4-1/2*3*4-1/2*12/5*16/5=154/25