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OpenStudy (anonymous):
In the figure, AC//DE, FG//BC and AD:DF:FB=1:2:3. If BE=10, find FG.
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OpenStudy (anonymous):
|dw:1418017855118:dw|
ganeshie8 (ganeshie8):
somethign wrong in the wording
ganeshie8 (ganeshie8):
`AD:DF=1:2:3` double check that part
OpenStudy (mathstudent55):
There is something missing from the ratio.
OpenStudy (anonymous):
haha corrected. thanks for reminding me
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OpenStudy (mathstudent55):
Is it AD:DF:FB?
OpenStudy (mathstudent55):
Oh I see.
OpenStudy (anonymous):
yup
OpenStudy (mathstudent55):
Numbers in circles represent ratios. Plain numbers are lengths of sides. |dw:1418017792347:dw|
OpenStudy (anonymous):
why FG is not 5?
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OpenStudy (mathstudent55):
|dw:1418017930012:dw|
ganeshie8 (ganeshie8):
FG = BC/2 not BE/2
OpenStudy (anonymous):
oh..
OpenStudy (mathstudent55):
FH/BE = DF/(DF + FB)
ganeshie8 (ganeshie8):
maybe find EC and use it
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OpenStudy (mathstudent55):
FH/10 = 2/5 FH = 4
OpenStudy (mathstudent55):
|dw:1418018182104:dw|
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