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shubhamsrg Group Title

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

  • 2 years ago
  • 2 years ago

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

    • 2 years ago
  2. shubhamsrg Group Title
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    |dw:1350050559015:dw|

    • 2 years ago
  3. CliffSedge Group Title
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    Ah.

    • 2 years ago
  4. CliffSedge Group Title
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    I can start by stating the obvious: SX/QX = 2/3.

    • 2 years ago
  5. shubhamsrg Group Title
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    hmm,,yep.. here's what i tried to do.. |dw:1350051162014:dw|

    • 2 years ago
  6. shubhamsrg Group Title
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    will this be helpful ?

    • 2 years ago
  7. CliffSedge Group Title
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    YM // to XZ? Hmm, yes, possibly..

    • 2 years ago
  8. shubhamsrg Group Title
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    yep.. //

    • 2 years ago
  9. shubhamsrg Group Title
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    further,, we may also see.. |dw:1350051445715:dw| let PZ = 2l thenZQ = 3l let ZM = j then MQ = 3l -j

    • 2 years ago
  10. shubhamsrg Group Title
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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 ?

    • 2 years ago
  11. sauravshakya Group Title
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    Is it 1/10

    • 2 years ago
  12. shubhamsrg Group Title
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    nopes..

    • 2 years ago
  13. shubhamsrg Group Title
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    nopes according to my book i mean,,i dont know! :P what was your solution ?

    • 2 years ago
  14. ganeshie8 Group Title
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    is it 1/3 ?

    • 2 years ago
  15. estudier Group Title
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    It's a/5-a whatever that is...:-)

    • 2 years ago
  16. shubhamsrg Group Title
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    nopes..not that either..

    • 2 years ago
  17. CliffSedge Group Title
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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

    • 2 years ago
  18. shubhamsrg Group Title
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    XQ/ZQ = SX/PZ ?? how sir ?

    • 2 years ago
  19. CliffSedge Group Title
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    Similar triangles.

    • 2 years ago
  20. shubhamsrg Group Title
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    ohh.lol..sorry,,i didnt compare with figure properly that one..hmm..

    • 2 years ago
  21. shubhamsrg Group Title
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    XQ/ZQ = 2/3 -> XQ = 2 will that be right to conclude ?

    • 2 years ago
  22. shubhamsrg Group Title
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    am saying this since this would be another assumption based on an earlier assumtion that PZ=2 and ZQ = 3

    • 2 years ago
  23. CliffSedge Group Title
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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..

    • 2 years ago
  24. CliffSedge Group Title
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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.

    • 2 years ago
  25. shubhamsrg Group Title
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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 ?

    • 2 years ago
  26. sauravshakya Group Title
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    Is it 2/5

    • 2 years ago
  27. estudier Group Title
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    SY/YQ should be 5/2 I think

    • 2 years ago
  28. estudier Group Title
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    Oops no, 5/3

    • 2 years ago
  29. CliffSedge Group Title
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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.

    • 2 years ago
  30. shubhamsrg Group Title
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    thats what am saying,,but you took both m and n = 1 ..

    • 2 years ago
  31. shubhamsrg Group Title
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    neither of those @sauravshakya and @estudier ans is 9/40

    • 2 years ago
  32. CliffSedge Group Title
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    I never bothered with m or n at all. I set PZ=2 and ZQ=3 and the rest followed.

    • 2 years ago
  33. ganeshie8 Group Title
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    gotcha !

    • 2 years ago
  34. shubhamsrg Group Title
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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..

    • 2 years ago
  35. shubhamsrg Group Title
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    @ganeshie8 am quite eager to know..

    • 2 years ago
  36. ganeshie8 Group Title
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    \(\triangle YZX \sim \triangle YRQ\)

    • 2 years ago
  37. shubhamsrg Group Title
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    oh,,i never really saw that,,cool !

    • 2 years ago
  38. estudier Group Title
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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)

    • 2 years ago
  39. ganeshie8 Group Title
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    and, \(\frac{ZQ}{PQ} = \frac{3}{5}\)

    • 2 years ago
  40. ganeshie8 Group Title
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    those two are sufficient to get to 9/40

    • 2 years ago
  41. shubhamsrg Group Title
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    am sorry but am still unable to @ganeshie8 :|

    • 2 years ago
  42. CliffSedge Group Title
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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.

    • 2 years ago
  43. shubhamsrg Group Title
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    ohh,,wait,,leme try again @ganeshie8

    • 2 years ago
  44. ganeshie8 Group Title
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    ok else id post... :) im seeing estudier posts.. .

    • 2 years ago
  45. shubhamsrg Group Title
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    thats really amazing.. gotcha @ganeshie8 @estudier @CliffSedge thanks all..really really great !!

    • 2 years ago
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