## mathmate Group Title 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. 2 years ago 2 years ago

1. mathmate Group Title

2. amistre64 Group Title

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 Group Title

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 Group Title

6x+4y+2z=12 12 12 12 /6 /4 /2 ------------- 2 3 6 are our intercepts

5. amistre64 Group Title

one way is we can take these and form vectors to get an angle to do a sin area formula with

6. arijit.mech Group Title

3.164 squre unit approx using area = squrt{ s(s-a)(s-b)(s-c)) ;s=(a+b+c)/2

7. amistre64 Group Title

(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 Group Title

So far I see Heron's formula from Amistre64 and arijit, and vectors from Amistre64. Please provide detailed calculations and answers.

9. amistre64 Group Title

my error is in the sqrt(58) :) sqrt(4+9) * sqrt(9+36) sqrt(13*45) 3 sqrt(65)

10. amistre64 Group Title

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 Group Title

12. amistre64 Group Title

i got a few more errors, but the concepts seems solid ;)

13. mathmate Group Title

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

14. amistre64 Group Title

right, and you can heron that for sheer simplicity

15. mathmate Group Title

Some "exact" answers would be nice! :)

16. mathmate Group Title

Agree, we also want as many different ways as possible.

17. amistre64 Group Title

|dw:1324567464017:dw|

18. amistre64 Group Title

sqrt(45) sin(y) = height sqrt(13) = base

19. amistre64 Group Title

b*h/2 = 3sqrt(14) if i did it right that time

20. mathmate Group Title

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 Group Title

the bh/2 is after the vectors give us a sin for an angle to determine the height

22. amistre64 Group Title

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 Group Title

$A= \frac{1}{2}\sqrt{|x|^2|y|^2-(x \cdot y)^2}$ 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 Group Title

work in progress here |dw:1324568575430:dw| $\int_{0}^{6}distance.x.to.y\ dz$

25. amistre64 Group Title

$X_i=\left(\frac{z-6}{3},0,z\right)$ $Y_i=\left(0,\frac{z-6}{2},z\right)$ $X_i-Y_i=\left(\frac{z-6}{3},\frac{-z+6}{2},0\right)$ 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 Group Title

hmm, wolf says that about 19, which is a bit off from the 3sqrt(14) i wonder why

27. amistre64 Group Title

348 not 468 ... but still thats off

28. amistre64 Group Title

oh well, it was a thought. guess i dont know how to freehand integrals yet :)

29. mathmate Group Title

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