点P1分线段AB 为5比7两部分,点P2分线段为5比11两部分,已知P1P2
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它的意思就是AP;PB=2:1,
∵点C把线段AB分为了9:5两部分,∴9+5=14.∵点O为线段AB的中点,∴14÷2=7cm.∴14÷14×9=9cm∴9-7=2cm∴线段OC长度=2cm.
设线段AB为x厘米5/(5+7)x-5/(5+11)x=10(1/12-1/16)x=2x=96AB长96cm
因为cd=ac-ad=5/12ab-5/16ab=5/48ab所以ab=48/5cd=48/5*10=96cm
10/(11/(11+5)-7/(5+7))=9610/(7/(5+7)-5/(5+11))
∵c点分线段ab为5:7两部分,D点分线段AB为5:1两部分∴AC=5/12ABAD=5/6AB∴CD=AD-AC=5/12AB=10∴AB=10÷5/12=24㎝再答:请采纳
由题设知:AC:CB=3:5.(AC+CB):CB=(3+5):5.[合比定理]AB:CB=8:5CB=(5/8)AB;又,AD:DB=7:5(AD+DB):DB=(7+5):5AB:DB=12:5D
∵C点分线段AB为3比5D分线段AB为3比7∴AC:BC=3:5AD:BD=3:7又∵AB=40∴AC=15BD=40*7/10=28∴DC=BD+AC-AB=15+28-40=3
AP1=5/(5+7)*AB=5/12*96=40(cm)AP2=5/(5+11)*AB=5/16*96=30(cm)P1P2=AP1-AP2=40-30=10(cm)
回答:(1)5/12-5/16=5/48;20cm/(5/48)=192cm(2)对折六次:10条
AC=5/12AB,AD=5/16AB,CD=2,CD=(5/12-5/16)AB=2,AB=2/(5/48)=19.6
设AB长为x(5/12-5/16)x=10解得x=96
设:AB=x,则AO=(5/12)x,BO=(7/12)x,AP=(5/16)x,PB=(11/16)xAP+PO+BO=AB(5/16)x+10+(7/12)x=x解得x=96即AB=96答:AB长
∵AC;BC=5:7∴AC=AB×5/(5+7)=5/12AB∵AD:BD=5:11∴AD=AB×5/(5+11)=5/16AB∴CD=AC-AD=5/12AB-5/16AB=5/48AB∵CD=15
/>∵AC:BC=5:7∴AC=5/(5+7)×AB=5/12AB∵AD:BD=5:11∴AD=5/(5+11)×AB=5/16AB∴CD=AC-AD=5/12AB-5/16AB=(20-15)/48
设AD=aDC=b=10CB=cAB=dd=a+10+c(a+10)/c=5/7a/(10+c)=5/11a=30b=10c=56AB=d=a+b+c=30+10+56=96
点C分线段AB5:7∴AC=5/12AB点D分线段AB为5:11∴AD=5/16AB∵CD=AC-AD=5㎝∴5/12AB-5/16AB=5㎝AB=48㎝
AC=5/12ABAD=5/16ABCD=AC-AD =(5/12-5/16)AB =5/48AB =5则,5/48
A——D—C——B∵AC:BC=5:7∴AC=5/(5+7)×AB=5/12AB∵AD:BD=5:11∴AD=5/(5+11)×AB=5/16AB∴CD=AC-AD=5/12AB-5/16AB=(20