shubhamsrg
  • shubhamsrg
geometry + similarity question i'd guess.. PQRS is a //gm PS//ZX PZ / QZ = 2/3 find XY/SQ =?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Is there a second parallelogram adjacent to the fist? I'm trying to imagine the scenario.
shubhamsrg
  • shubhamsrg
|dw:1350050559015:dw|
anonymous
  • anonymous
Ah.

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

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