作业帮初一数学第四章有关线段计算20题及答案
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初一数学第四章有关线段计算20题及答案
延长线段AB C,BC = 2AB,D是中点的线段AB,E,F是BC上的一点,和BE：EF = 1:2,EF：FG = 2：5,AC = 60CM ,求DE DE长的答案：已知BC = 2AB,DB = 1/2AB,交流=60厘米
到DB = 1/4BC = 1/6AC = 10厘米.
所以BC = 40厘米.
因为BE：EF = 1:2,EF：FC = 2:5,BC = BE + EF + FC
BC = 2 +4 / 5BE = 19/5BE /> = 200/19cm.
EF = 2BE = 400/19cm
所以DE = DB + BE = 10 +200 / 19 = 390/19cm
DF = DE + EF = 390/19 +400 / 19 = 790 /19厘米
已知线段AB =10厘米,C是AB的中点,AC和DC：DB = 2点07分上的点,求线段AD的长度
答：有两种可能的
a>
___________________
ADCB
成立,CD = 2A,DB = 7A
BC = DB-CD = 7A-2A = 5A
另一个尹蔚AB = 10
所以BC = 5
所以5A = 5
= 1
AD = 10-DB = 10-7A = 3
___________________
ACDB
10/2 = 5
2 +7 = 9
CD = 5 * 2/9 = 10/9
AD = 10 / 9 +5 = 55/9
到DB = 1/4BC = 1/6AC = 10厘米.
所以BC = 40厘米.
因为BE：EF = 1:2,EF：FC = 2:5,BC = BE + EF + FC
BC = 2 +4 / 5BE = 19/5BE /> = 200/19cm.
EF = 2BE = 400/19cm
所以DE = DB + BE = 10 +200 / 19 = 390/19cm
DF = DE + EF = 390/19 +400 / 19 = 790 /19厘米
已知线段AB =10厘米,C是AB的中点,AC和DC：DB = 2点07分上的点,求线段AD的长度
答：有两种可能的
a>
___________________
ADCB
成立,CD = 2A,DB = 7A
BC = DB-CD = 7A-2A = 5A
另一个尹蔚AB = 10
所以BC = 5
所以5A = 5
= 1
AD = 10-DB = 10-7A = 3
___________________
ACDB
10/2 = 5
2 +7 = 9
CD = 5 * 2/9 = 10/9
AD = 10 / 9 +5 = 55/9