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estudier

  • 3 years ago

Area of square?

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  1. estudier
    • 3 years ago
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    |dw:1350208678915:dw|

  2. sauravshakya
    • 3 years ago
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    What is that?

  3. Jonask
    • 3 years ago
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    is that 5,7,10

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

  5. estudier
    • 3 years ago
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    9,10,7

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

  7. experimentX
    • 3 years ago
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    haha ... you are kidding!!

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

  9. experimentX
    • 3 years ago
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    is that all info??

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

  11. Jonask
    • 3 years ago
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    and the side is less than 17

  12. Jonask
    • 3 years ago
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    according to triangle equality

  13. experimentX
    • 3 years ago
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    |dw:1350209046775:dw|

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

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

  16. hartnn
    • 3 years ago
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    |dw:1350209083317:dw|

  17. Jonask
    • 3 years ago
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    i meant 7^2|dw:1350176811646:dw|

  18. estudier
    • 3 years ago
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    JonasK is on the right track

  19. Jonask
    • 3 years ago
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    what about using the area rule

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

  21. estudier
    • 3 years ago
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    Probably...

  22. estudier
    • 3 years ago
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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

  23. Jonask
    • 3 years ago
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    yes

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

  25. Jonask
    • 3 years ago
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    okay

  26. estudier
    • 3 years ago
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    Cos of angle opposite 7 is sin theta....

  27. experimentX
    • 3 years ago
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    |dw:1350209708034:dw|

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

  29. estudier
    • 3 years ago
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    As JonasK had it at first...

  30. estudier
    • 3 years ago
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    expX, now eliminate theta....

  31. experimentX
    • 3 years ago
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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

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

  33. estudier
    • 3 years ago
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    Don't be lazy....:-)

  34. experimentX
    • 3 years ago
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    haha ..

  35. estudier
    • 3 years ago
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    Sin and Cos should auto response.....

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

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

  38. Jonask
    • 3 years ago
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    with - before

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

  40. experimentX
    • 3 years ago
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    haa ... i was being fool trying to eliminate theta directly.

  41. experimentX
    • 3 years ago
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    |dw:1350210145598:dw|

  42. estudier
    • 3 years ago
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    A bit messy but sufficient...:-)

  43. Jonask
    • 3 years ago
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    seems great

  44. experimentX
    • 3 years ago
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    still quartic equation is pretty awful ... i must say!!

  45. estudier
    • 3 years ago
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    Yes, squaring and adding gets shot of theta...

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

  47. estudier
    • 3 years ago
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    Which is quadratic in x^2

  48. experimentX
    • 3 years ago
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    Oh .. that's pretty nice!!

  49. experimentX
    • 3 years ago
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    direct answer.

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

  51. Jonask
    • 3 years ago
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    @estudier seems better

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

  53. shubhamsrg
    • 3 years ago
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    correction : x+2y = 10

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

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

  56. Jonask
    • 3 years ago
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    not sure

  57. estudier
    • 3 years ago
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    Eh? Some extra ssumptions there....

  58. shubhamsrg
    • 3 years ago
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    ohh..lol..so sorry! :P

  59. Jonask
    • 3 years ago
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    no prob

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

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

  62. estudier
    • 3 years ago
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    65 + 14 sqrt 14 I think.

  63. Jonask
    • 3 years ago
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    same thing i just approaximated it

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

  65. Jonask
    • 3 years ago
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    \[A=65+14\sqrt{14}\]

  66. Jonask
    • 3 years ago
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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,

  67. shubhamsrg
    • 3 years ago
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    equilateral ?

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

  69. shubhamsrg
    • 3 years ago
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    you mean like this: |dw:1350211506937:dw| ??

  70. Jonask
    • 3 years ago
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    yes

  71. shubhamsrg
    • 3 years ago
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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..

  72. sauravshakya
    • 3 years ago
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  73. Jonask
    • 3 years ago
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    nice work @sauravshakya

  74. sauravshakya
    • 3 years ago
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    Thanx @Jonask

  75. Jonask
    • 3 years ago
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    what disqualifies 65-14root14

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

  77. sauravshakya
    • 3 years ago
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    I think I have written reason for that too.

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

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