Question

E.F分别为矩形ABCD的BC边和CD边上的点.且S△ABE=2,S△ECF=3,S△ADF=4

Answer 1

S△ABE=2=AB*BE/2,S△ECF=3=EC*CF/2,S△ADF=4=AD*DF/2

设AB=X,AD=Y;

则BE=4/X;DF=8/Y;

EC*CF=(AD-BE)(AB-DF)=(Y-4/X)(X-8/Y)=6;

(XY)^2-18XY+32=0;

因为S(ABCD）=XY,所以XY=16,得S△AEF=16-2-3-4=7