shubhamsrg
geometry + similarity question i'd guess..
PQRS is a //gm
PS//ZX
PZ / QZ = 2/3
find XY/SQ =?
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CliffSedge
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Is there a second parallelogram adjacent to the fist? I'm trying to imagine the scenario.
shubhamsrg
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|dw:1350050559015:dw|
CliffSedge
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Ah.
CliffSedge
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I can start by stating the obvious: SX/QX = 2/3.
shubhamsrg
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hmm,,yep..
here's what i tried to do..
|dw:1350051162014:dw|
shubhamsrg
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will this be helpful ?
CliffSedge
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YM // to XZ?
Hmm, yes, possibly..
shubhamsrg
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yep.. //
shubhamsrg
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further,, we may also see..
|dw:1350051445715:dw|
let PZ = 2l
thenZQ = 3l
let ZM = j
then MQ = 3l -j
shubhamsrg
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|dw:1350051539893:dw|
we can easily see
a/(a+b+c) = 2/5
b/(a+b+c)= j/5l
c/(a+b+c) =(3l -j)/5l
we need to find numerical value of b/(a+b+c)
where, we know that
a/(b+c) = 2/3
any help ?
sauravshakya
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Is it 1/10
shubhamsrg
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nopes..
shubhamsrg
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nopes according to my book i mean,,i dont know! :P
what was your solution ?
ganeshie8
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is it 1/3 ?
estudier
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It's a/5-a whatever that is...:-)
shubhamsrg
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nopes..not that either..
CliffSedge
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Here's what I got so far.
I let PZ=2 and ZQ=3 since that satisfies the ratio.
XQ/ZQ = SX/PZ
-> XQ/3 = SX/2
XQ/ZQ = 2/3
-> XQ = 2
SX/PZ = XQ/ZQ
-> SX/2 = 2/3
SX = 4/3
SX+XQ = 10/3
shubhamsrg
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XQ/ZQ = SX/PZ ?? how sir ?
CliffSedge
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Similar triangles.
shubhamsrg
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ohh.lol..sorry,,i didnt compare with figure properly that one..hmm..
shubhamsrg
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XQ/ZQ = 2/3
-> XQ = 2
will that be right to conclude ?
shubhamsrg
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am saying this since this would be another assumption based on an earlier assumtion that PZ=2 and ZQ = 3
CliffSedge
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I think so, that's what I got.
I'm still playing around with the proportions to find XY or YQ..
CliffSedge
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That wasn't an assumption, that was an arbitrary choice. There's no restriction on what PZ and ZQ are as long as they satisfy the ratio.
shubhamsrg
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but, in generailty, we have PZ = 2m and ZQ = 3m
you assumed m=1 ,,fair enough
now XQ/ZQ = 2/3
XQ = 2n => ZQ = 3n
this came from the fact m =1//
now will that be right in saying n=1 also when m=1 already ?
sauravshakya
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Is it 2/5
estudier
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SY/YQ should be 5/2 I think
estudier
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Oops no, 5/3
CliffSedge
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If you're using 2m and 3m and 2n and 3n, just because you can choose m=1 at first, that doesn't make n=1.
shubhamsrg
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thats what am saying,,but you took both m and n = 1 ..
shubhamsrg
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neither of those @sauravshakya and @estudier
ans is 9/40
CliffSedge
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I never bothered with m or n at all. I set PZ=2 and ZQ=3 and the rest followed.
ganeshie8
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gotcha !
shubhamsrg
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PZ=2 and ZQ = 3 made you reach till
XQ/ZQ = 2/3
but XQ = 2 and ZQ=3 i guess wont be right..
shubhamsrg
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@ganeshie8 am quite eager to know..
ganeshie8
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\(\triangle YZX \sim \triangle YRQ\)
shubhamsrg
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oh,,i never really saw that,,cool !
estudier
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I wasn't saying 5/3 was the answer, I was saying that if RZ divides PQ in ratio 2 to 3 then Y divides SQ in ratio 2+3 to 3 (imagine RZ sliding along QP until it becomes a diagonal of the //gram)
ganeshie8
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and, \(\frac{ZQ}{PQ} = \frac{3}{5}\)
ganeshie8
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those two are sufficient to get to 9/40
shubhamsrg
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am sorry but am still unable to @ganeshie8 :|
CliffSedge
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I was able to get it using estudier's SY/YQ=5/3.
SY + YQ = SQ = 10/3
SY = 10/3 - YQ
(10/3-/YQ)/YQ = 5/3
-> YQ=5/4
SY = 25/12
XY = SY-SX
-> XY = 3/4
XY/SQ = (3/4)/(10/3) = 9/40.
QED.
shubhamsrg
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ohh,,wait,,leme try again @ganeshie8
ganeshie8
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ok else id post... :) im seeing estudier posts.. .
shubhamsrg
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thats really amazing..
gotcha @ganeshie8 @estudier @CliffSedge
thanks all..really really great !!