AE:ED=AF:AB=BG:GC=k,把这个表达式变换成k和矩形ABCD边长a、b的表达式,
则有:AE=BG=kb:(k+1)
ED=GC=
b |
k+1 |
AF=ka,FB=(1-k)a
S(矩形ABCD)=ab=S(Rt△AFE)+S(△FEC)+S( Rt△EDC)+S(Rt△FBC)
=
1 |
2 |
1 |
2 |
1 |
2 |
=
1 |
2 |
1 |
2 |
1 |
2 |
=
1 |
k+1 |
解ab,得:
ab=
20×(k+1) |
k |
同理S(矩形ABCD)=ab=S(Rt△FBG)+S(△FGD)+S( Rt△GDC)+S(Rt△AFD)
=
1 |
2 |
1 |
2 |
1 |
2 |
=
1 |
2 |
k+1 |
kb |
1 |
2 |
b |
k+1 |
1 |
2 |
=
2k+1 |
2k+2 |
解ab,得:
ab=32(k+1)(2)
根据(1)(2),解得k=
5 |
8 |