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mathmate

  • 2 years ago

QUIZ TIME : easy question : area of triangle The plane 6x+4y+2z=12 intersects the coordinate planes to form three sides of a triangle. Find the area of the triangle. The challenge here is to show as many ways as you can find to get the correct answer. Give exact values if you can, eg. sqrt(14) instead of 3.742. NOTE to readers: Please award a medal to someone who presents a formula or method that you would not have used.

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  1. mathmate
    • 2 years ago
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  2. amistre64
    • 2 years ago
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    x=0, get the intercepts; put them in a distance formula .... di this for each variable to zero out. then you have the side measures and can do a heron on it

  3. amistre64
    • 2 years ago
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    or... divide by 12 and divide off the tops to get the intercepts if we dont have time to do them one by one

  4. amistre64
    • 2 years ago
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    6x+4y+2z=12 12 12 12 /6 /4 /2 ------------- 2 3 6 are our intercepts

  5. amistre64
    • 2 years ago
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    one way is we can take these and form vectors to get an angle to do a sin area formula with

  6. arijit.mech
    • 2 years ago
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    3.164 squre unit approx using area = squrt{ s(s-a)(s-b)(s-c)) ;s=(a+b+c)/2

  7. amistre64
    • 2 years ago
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    (2,0,0) (0,0,6) -(0,3,0) -(0,3,0) ------- ------- <2,-3,0> <0,-3,6> cos(yaxis) = <2,-3,0>.<0,-3,6> 9 ----------------- = --------------- |<2,-3,0>| |<0,-3,6>| sqrt(4+9+9+36) cos(y) = 9/sqrt(58) y = cos-1(9/sqrt(58)) |<2,-3,0>| |<0,-3,6>| sin(cos-1(9/sqrt(58))) / 2 9/sqrt(58) sin(cos-1(9/sqrt(58))) / 2 maybe :)

  8. mathmate
    • 2 years ago
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    So far I see Heron's formula from Amistre64 and arijit, and vectors from Amistre64. Please provide detailed calculations and answers.

  9. amistre64
    • 2 years ago
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    my error is in the sqrt(58) :) sqrt(4+9) * sqrt(9+36) sqrt(13*45) 3 sqrt(65)

  10. amistre64
    • 2 years ago
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    Base * (..... ......height..........) /2 9/3sqrt(65) sin(cos-1(9/3sqrt(65))) / 2 3/sqrt(65) sin(cos-1(3/sqrt(65))) / 2 hopefully lol

  11. arijit.mech
    • 2 years ago
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    PLEASE SEE THE DRAWING........

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  12. amistre64
    • 2 years ago
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    i got a few more errors, but the concepts seems solid ;)

  13. mathmate
    • 2 years ago
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    So 2,3,6 are the intercepts, sqrt(13), sqrt(45), sqrt(40) are the sides.

  14. amistre64
    • 2 years ago
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    right, and you can heron that for sheer simplicity

  15. mathmate
    • 2 years ago
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    Some "exact" answers would be nice! :)

  16. mathmate
    • 2 years ago
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    Agree, we also want as many different ways as possible.

  17. amistre64
    • 2 years ago
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    |dw:1324567464017:dw|

  18. amistre64
    • 2 years ago
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    sqrt(45) sin(y) = height sqrt(13) = base

  19. amistre64
    • 2 years ago
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    b*h/2 = 3sqrt(14) if i did it right that time

  20. mathmate
    • 2 years ago
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    Wow! First exact answer! 3sqrt(14) (that's what I have too). So far we have intercepts: 2,3,6 sides : sqrt(13), sqrt(45), sqrt(40) Area : 3sqrt(14) Methods: Heron : Amistre64, arigit Vectors: Amistre64 (to be completed) bh/2 : Amistre64

  21. amistre64
    • 2 years ago
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    the bh/2 is after the vectors give us a sin for an angle to determine the height

  22. amistre64
    • 2 years ago
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    i wonder if its worth it to do a 3d integration :) or at least take the measures and translate them to a xy congruent triangle

  23. phi
    • 2 years ago
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    \[A= \frac{1}{2}\sqrt{|x|^2|y|^2-(x \cdot y)^2}\] with (using amistre's two vectors) x=<2,-3,0> y= <0,-3,6> gives 0.5sqrt( 13*45-81)= 1.5sqrt(65-9)= 1.5sqrt(56)= 3sqrt(14)

  24. amistre64
    • 2 years ago
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    work in progress here |dw:1324568575430:dw| \[\int_{0}^{6}distance.x.to.y\ dz\]

  25. amistre64
    • 2 years ago
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    \[X_i=\left(\frac{z-6}{3},0,z\right)\] \[Y_i=\left(0,\frac{z-6}{2},z\right)\] \[X_i-Y_i=\left(\frac{z-6}{3},\frac{-z+6}{2},0\right)\] distance from X to Y:\[\sqrt{(\frac{z-6}{3})^2+(\frac{-z+6}{2})^2}\] \[\sqrt{\frac{z^2-6z+36}{9}+\frac{z^2-6z+36}{4}}\] \[\sqrt{\frac{4z^2-4.6z+4.36+9z^2-9.6z+9.36}{36}}\] \[\frac{\sqrt{13z^2-78z+468}}{6}\] ergo lol \[\int_{0}^{6} \frac{\sqrt{13z^2-78z+468}}{6}dz\]

  26. amistre64
    • 2 years ago
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    hmm, wolf says that about 19, which is a bit off from the 3sqrt(14) i wonder why

  27. amistre64
    • 2 years ago
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    348 not 468 ... but still thats off

  28. amistre64
    • 2 years ago
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    oh well, it was a thought. guess i dont know how to freehand integrals yet :)

  29. mathmate
    • 2 years ago
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    OK, here's what we have so far (let me know if I missed anything) Phi: does the formula come from cross product, it's neat, like a generalized form of bh/2. Related to that, I suggest the cross product of Amistre64's two vectors. (1/2)AxB=(1/2)<0,-3,6>x<2,-3,0>=(1/2)sqrt(504)=3sqrt(14) Methods: 1. Heron : Amistre64, arigit 2. (1/2)xysin(theta) + bh/2 : Amistre64 3. integration : Amistre64 (in progress) 4. (1/2)sqrt(|x|^2|y|^2-x.y) : phi 5. half magnitude of cross-product, (1/2)|PxQ|: mathmate

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