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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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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
or... divide by 12 and divide off the tops to get the intercepts if we dont have time to do them one by one

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Other answers:

6x+4y+2z=12 12 12 12 /6 /4 /2 ------------- 2 3 6 are our intercepts
one way is we can take these and form vectors to get an angle to do a sin area formula with
3.164 squre unit approx using area = squrt{ s(s-a)(s-b)(s-c)) ;s=(a+b+c)/2
(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 :)
So far I see Heron's formula from Amistre64 and arijit, and vectors from Amistre64. Please provide detailed calculations and answers.
my error is in the sqrt(58) :) sqrt(4+9) * sqrt(9+36) sqrt(13*45) 3 sqrt(65)
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
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i got a few more errors, but the concepts seems solid ;)
So 2,3,6 are the intercepts, sqrt(13), sqrt(45), sqrt(40) are the sides.
right, and you can heron that for sheer simplicity
Some "exact" answers would be nice! :)
Agree, we also want as many different ways as possible.
sqrt(45) sin(y) = height sqrt(13) = base
b*h/2 = 3sqrt(14) if i did it right that time
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
the bh/2 is after the vectors give us a sin for an angle to determine the height
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
  • phi
\[A= \frac{1}{2}\sqrt{|x|^2|y|^2-(x \cdot y)^2}\] 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)
work in progress here |dw:1324568575430:dw| \[\int_{0}^{6}\ dz\]
\[X_i=\left(\frac{z-6}{3},0,z\right)\] \[Y_i=\left(0,\frac{z-6}{2},z\right)\] \[X_i-Y_i=\left(\frac{z-6}{3},\frac{-z+6}{2},0\right)\] 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 \[\int_{0}^{6} \frac{\sqrt{13z^2-78z+468}}{6}dz\]
hmm, wolf says that about 19, which is a bit off from the 3sqrt(14) i wonder why
348 not 468 ... but still thats off
oh well, it was a thought. guess i dont know how to freehand integrals yet :)
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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