Maths......
回答 (3)
2007-04-04 4:11 am
✔ 最佳答案
DE//BD=>△KDE~△KCB
=>KD:KC=KE:BK}
KD=2,KE=3,KC=5,KB=b}
=>b=(3*5)/2=7.5
DE//BD
=>△KDE~△KCB
=>KD:KC=DE:BC}
KD=2,KC=5,DE=4,BC=a}
=>a=(5*4)/2=10
DE//BD
=>△ADE~△ABD
=>AD:AB=ED:BC}
AD=5,AB=(5+c),ED=4,BC=a=10}
=>5+c=(5*10)/4=12.5
=>c=7.5
2007-04-04 4:09 am
angle deb=ebc(de parallel to bc)
angle edk=kcb(de parallel to bc)
angle dke= bkc(vert. opp angle)
therefore triangle deb similar to ebc(aaa)
cb/dk = ed/kc
a/2 = 4/5
a=2.5
bk/ek=ed/kc
b/3=4/5
b=2.4
c唔記得點做
angle edk=kcb(de parallel to bc)
angle dke= bkc(vert. opp angle)
therefore triangle deb similar to ebc(aaa)
cb/dk = ed/kc
a/2 = 4/5
a=2.5
bk/ek=ed/kc
b/3=4/5
b=2.4
c唔記得點做
2007-04-04 4:06 am
a/4=5/2
a=10
b/3=5/2
b=7.5
4/a=5/(5+c)
4/10=5/(5+c)
c=7.5
a=10
b/3=5/2
b=7.5
4/a=5/(5+c)
4/10=5/(5+c)
c=7.5
收錄日期: 2021-04-12 23:49:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070403000051KK03778