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estudier

Area of square?

  • one year ago
  • one year ago

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  1. estudier
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    |dw:1350208678915:dw|

    • one year ago
  2. sauravshakya
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    What is that?

    • one year ago
  3. Jonask
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    is that 5,7,10

    • one year ago
  4. estudier
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    A square, a point inside of it distances from 3 vertices as shown

    • one year ago
  5. estudier
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    9,10,7

    • one year ago
  6. estudier
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    This is for experimentX, who loves algebra..:-)

    • one year ago
  7. experimentX
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    haha ... you are kidding!!

    • one year ago
  8. estudier
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    It's not too hard..(it's still too early)

    • one year ago
  9. experimentX
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    is that all info??

    • one year ago
  10. Jonask
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    if we call ne side a then the area is \[A=a^2\]

    • one year ago
  11. Jonask
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    and the side is less than 17

    • one year ago
  12. Jonask
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    according to triangle equality

    • one year ago
  13. experimentX
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    |dw:1350209046775:dw|

    • one year ago
  14. Jonask
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    \[9^2=a^2+10^2-2(10)(a)\cos \theta\] \[762=10^2+a^2-2(a)(10)\cos (90-\theta)\]

    • one year ago
  15. estudier
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    2 var 1 equation suggests simultaneous equations as the way to go....

    • one year ago
  16. hartnn
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    |dw:1350209083317:dw|

    • one year ago
  17. Jonask
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    i meant 7^2|dw:1350176811646:dw|

    • one year ago
  18. estudier
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    JonasK is on the right track

    • one year ago
  19. Jonask
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    what about using the area rule

    • one year ago
  20. experimentX
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    probably @hartnn method would work out too ... since they are all independent equations.

    • one year ago
  21. estudier
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    Probably...

    • one year ago
  22. estudier
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    "what about using the area rule" U just want to get rid of the 90- bit and then you can get rid of theta

    • one year ago
  23. Jonask
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    yes

    • one year ago
  24. experimentX
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    still that's quite a bit of algebra since they are all non linear equations ... let's try something better.

    • one year ago
  25. Jonask
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    okay

    • one year ago
  26. estudier
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    Cos of angle opposite 7 is sin theta....

    • one year ago
  27. experimentX
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    |dw:1350209708034:dw|

    • one year ago
  28. estudier
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    What I mean is that if angle opposite 9 is theta, then cos of angle opposite 7 is sin theta

    • one year ago
  29. estudier
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    As JonasK had it at first...

    • one year ago
  30. estudier
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    expX, now eliminate theta....

    • one year ago
  31. experimentX
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    you know i rely on wolf for this http://www.wolframalpha.com/input/?i=2*10*a+sin%28theta%29+%3D+10^2+%2B+a^2++-+9^2%2C+2*10*a+cos%28theta%29+%3D+10^2+%2B+a^2++-+7^2

    • one year ago
  32. Jonask
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    \[32=-20(a)\cos \theta+20(a)\sin \theta\]

    • one year ago
  33. estudier
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    Don't be lazy....:-)

    • one year ago
  34. experimentX
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    haha ..

    • one year ago
  35. estudier
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    Sin and Cos should auto response.....

    • one year ago
  36. Jonask
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    \[\frac{ 32 }{ -20 a}=\cos \theta -\sin \theta\]

    • one year ago
  37. Jonask
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    \[\cos \theta -\sin \theta =\sqrt{2} \sin(45-\theta)\]

    • one year ago
  38. Jonask
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    with - before

    • one year ago
  39. estudier
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    @Jonask An identity (well known) in Sin and Cos?

    • one year ago
  40. experimentX
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    haa ... i was being fool trying to eliminate theta directly.

    • one year ago
  41. experimentX
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    |dw:1350210145598:dw|

    • one year ago
  42. estudier
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    A bit messy but sufficient...:-)

    • one year ago
  43. Jonask
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    seems great

    • one year ago
  44. experimentX
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    still quartic equation is pretty awful ... i must say!!

    • one year ago
  45. estudier
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    Yes, squaring and adding gets shot of theta...

    • one year ago
  46. estudier
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    U get after simplification...400x^2 = (x^2 +19)^2 + (x^2 +51)^2 -> x^4 -130 x^2 +1481

    • one year ago
  47. estudier
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    Which is quadratic in x^2

    • one year ago
  48. experimentX
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    Oh .. that's pretty nice!!

    • one year ago
  49. experimentX
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    direct answer.

    • one year ago
  50. Jonask
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    let \[-10^2-a^2=b\] \[\frac{ (7^2 -b)^2}{ (-20a )^2}+\frac{ (9^2-b)^2 }{ (20a)^2 }=1\]

    • one year ago
  51. Jonask
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    @estudier seems better

    • one year ago
  52. shubhamsrg
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    |dw:1350210434675:dw| x+2y = 7 ..(1) y^2 + (x+y)^2 = 81 ..(2) hope that does it..

    • one year ago
  53. shubhamsrg
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    correction : x+2y = 10

    • one year ago
  54. shubhamsrg
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    solving shouldnt be much problem since from (1) , x+y = 10-y and we plug that directly in (2)

    • one year ago
  55. Jonask
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    the line with lenght 10 is not necessarily a diagonal when extrapolated ??

    • one year ago
  56. Jonask
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    not sure

    • one year ago
  57. estudier
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    Eh? Some extra ssumptions there....

    • one year ago
  58. shubhamsrg
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    ohh..lol..so sorry! :P

    • one year ago
  59. Jonask
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    no prob

    • one year ago
  60. Jonask
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    \[a=10,834, a=3,552\]

    • one year ago
  61. Jonask
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    can we eliminate according to triangle inequality 3,552 such that \[A=(10,834)^2\]

    • one year ago
  62. estudier
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    65 + 14 sqrt 14 I think.

    • one year ago
  63. Jonask
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    same thing i just approaximated it

    • one year ago
  64. Jonask
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    \[A=(\sqrt{64+14\sqrt{14}} )^2\]

    • one year ago
  65. Jonask
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    \[A=65+14\sqrt{14}\]

    • one year ago
  66. Jonask
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    @shubhamsrg if we consider the triangles to be equilateral then how wuld we solve it |dw:1350178935156:dw| with common side 10,

    • one year ago
  67. shubhamsrg
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    equilateral ?

    • one year ago
  68. Jonask
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    |dw:1350179036162:dw| such that the legth is now diagonal

    • one year ago
  69. shubhamsrg
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    you mean like this: |dw:1350211506937:dw| ??

    • one year ago
  70. Jonask
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    yes

    • one year ago
  71. shubhamsrg
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    for that i had given the solution above.. x+2y = 10 ..(1) y^2 + (x+y)^2 = 81 ..(2) from (1), x+y = 10-y and we plug this in (2) to solve for y, and x and ultimately area..

    • one year ago
  72. sauravshakya
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    • one year ago
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  73. Jonask
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    nice work @sauravshakya

    • one year ago
  74. sauravshakya
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    Thanx @Jonask

    • one year ago
  75. Jonask
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    what disqualifies 65-14root14

    • one year ago
  76. estudier
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    @Jonask That's for x^2, remember....

    • one year ago
  77. sauravshakya
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    I think I have written reason for that too.

    • one year ago
  78. estudier
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    @sauravshakya Yes, good job (I think he just forgot it was for x^2)

    • one year ago
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