Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

(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 :)

Answer 7

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

Answer 8

my error is in the sqrt(58) :)

sqrt(4+9) * sqrt(9+36)

sqrt(13*45)

3 sqrt(65)

Answer 9

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

Answer 10

PLEASE SEE THE DRAWING........

Answer 11

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

Answer 12

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

Answer 13

right, and you can heron that for sheer simplicity

Answer 14

Some "exact" answers would be nice! :)

Answer 15

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