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geometry + similarity question i'd guess.. PQRS is a //gm PS//ZX PZ / QZ = 2/3 find XY/SQ =?

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Is there a second parallelogram adjacent to the fist? I'm trying to imagine the scenario.

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I can start by stating the obvious: SX/QX = 2/3.
hmm,,yep.. here's what i tried to do.. |dw:1350051162014:dw|
will this be helpful ?
YM // to XZ? Hmm, yes, possibly..
yep.. //
further,, we may also see.. |dw:1350051445715:dw| let PZ = 2l thenZQ = 3l let ZM = j then MQ = 3l -j
|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 ?
Is it 1/10
nopes according to my book i mean,,i dont know! :P what was your solution ?
is it 1/3 ?
It's a/5-a whatever that is...:-)
nopes..not that either..
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
XQ/ZQ = SX/PZ ?? how sir ?
Similar triangles.,,i didnt compare with figure properly that one..hmm..
XQ/ZQ = 2/3 -> XQ = 2 will that be right to conclude ?
am saying this since this would be another assumption based on an earlier assumtion that PZ=2 and ZQ = 3
I think so, that's what I got. I'm still playing around with the proportions to find XY or YQ..
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.
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 ?
Is it 2/5
SY/YQ should be 5/2 I think
Oops no, 5/3
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.
thats what am saying,,but you took both m and n = 1 ..
neither of those @sauravshakya and @estudier ans is 9/40
I never bothered with m or n at all. I set PZ=2 and ZQ=3 and the rest followed.
gotcha !
PZ=2 and ZQ = 3 made you reach till XQ/ZQ = 2/3 but XQ = 2 and ZQ=3 i guess wont be right..
@ganeshie8 am quite eager to know..
\(\triangle YZX \sim \triangle YRQ\)
oh,,i never really saw that,,cool !
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)
and, \(\frac{ZQ}{PQ} = \frac{3}{5}\)
those two are sufficient to get to 9/40
am sorry but am still unable to @ganeshie8 :|
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.
ohh,,wait,,leme try again @ganeshie8
ok else id post... :) im seeing estudier posts.. .
thats really amazing.. gotcha @ganeshie8 @estudier @CliffSedge thanks all..really really great !!

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