http://hk.myblog.yahoo.com/kitlkw92/article?mid=5&new=1
第3戈題點計牙??
just第3喳..
要解釋
f.3 maths ---四邊形...急
2007-02-25 11:30 pm
回答 (4)
2007-02-25 11:44 pm
✔ 最佳答案
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
參考: me
2007-02-26 1:29 am
△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)
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)
參考: 自己
2007-02-25 11:56 pm
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
2007-02-25 11:54 pm
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)
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)
收錄日期: 2021-04-12 23:35:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070225000051KK03034