f.3 maths ---四邊形...急

triangle FEB ~ triangle ADF (AAA)

Let CE = a, BE = 8-a = b, a + b = 8
FD = CE = a = AD
FE = DC = 8 - AD = b

therefore
FE = DC = BE = b

FE + DC + EC + FD
= 2b + 2a
=2(a+b)
=2x8
=16

周界 = 16cm

△ABC ~ △FBE (AAA similarity)
所以△FEB是等腰直角三角形

FE = BE = 8-EC

長方形DCEF的周界
2 (FE+EC)
= 2 (8-EC+EC)
= 2 x 8
= 16 (cm^2)

因為ABC係等腰直角三角形，角BAC=FAD=45度，角ABC=FBE=45度。角BEF=180-90=90度，BEF係等腰直角三角形；角FDA=180-90=90度，FAD係等腰直角三角形。FE=BE=x cm，FD=AD=(8-x) cm，DCEF周界=(x+(8-x))*2=16cm

S△ABC=(8cm*8cm)/2=32cm^2}
S△ABC=S△BEF+S△AFD+S長方形FECD}

=>32=S△BEF+S△AFD+S長方形FECD}
S△BEF=[(8cm-EC)*EF]/2}
S△AFD=[(8cm-EF)*EC]/2}
S長方形FECD= EF*EC}

=>32=[(8-EC)*EF]/2+(8-EF)*EC]/2+ EF*EC
....64=8EF-EF*EC+8EC- EF*EC+ 2EF*EC
....64=8EF+8EC
....8= EF+EC

=>長方形DCEF的周界為2(EF+EC)=16(cm)