Question

What is the area of the parallelogram with vertices A(1,5,0), B(6,10,−3), C(−4,5,−2), and D(1,10,−5)?

Answer 1

`S_(ABCD) = 15*sqrt(6)~=36.742`

Explanation:

First we should verify that the 4 points really are coplanar. Following, for instance, the method described in:
Equation of a Plane determined by 3 points

I checked it obtaining the plane defined by the points A, B and C, then trying D in the equation of the plane, verifying that these 4 points are indeed coplanar.

Further: note that the figure ABCD is composed of the triangles ABC and ACD.
Repeating the coordinates of the points:
A(1,5,0)
B(6,10,-3)
C(-4,5,-2)
D(1,10,-5)

Obtaining sides of the triangles ABC and ACD:
`AB=sqrt((6-1)^2+(10-5)^2+(-3-0)^2)=sqrt(25+25+9)=sqrt(59)`
`AC=sqrt((-4-1)^2+(5-5)^2+(-2-0)^2)=sqrt(25+0+4)=sqrt(29)`
`BC=sqrt((-4-6)^2+(5-10)^2+(-2+3)^2)=sqrt(100+25+1)=sqrt(126)`
`CD=sqrt((1+4)^2+(10-5)^2+(-5+2)^2)=sqrt(25+25+9)=sqrt(59)`
`DA=sqrt((1-1)^2+(5-10)^2+(0+5)^2)=sqrt(0+25+25)=sqrt(50)`

Just a minute! `AB=CD` but `BC!=DA!` Let's calculate also BD:
`BD=sqrt((1-6)^2+(10-10)^2+(-5+3)^2)=sqrt(25+0+4)=sqrt(29)`

Since `AB=CD` and `AC=BD` the parallelogram should be called ABDC , composed of the triangles ABC and ABD, and, because these triangles are congruent, `S_(triangleABC) = S_(triangleBDC)` and `S_("parallelogram" ABDC) = 2*S_(triangleABC)`

Using Heron's Formula
Triangle ABC (`a=AB, b=AC and c=BC`)
`s=(a+b+c)/2=(sqrt(59)+sqrt(29)+sqrt(126))/2~=12.14564`
`(s-a)=(-sqrt(59)+sqrt(29)+sqrt(126))/2~=4.46450`
`(s-b)=(sqrt(59)-sqrt(29)+sqrt(126))/2~=6.76048`
`(s-c)=(sqrt(59)+sqrt(29)-sqrt(126))/2~=0.92067`
`S_(triangleABC)=sqrt(s(s-a)(s-b)(s-c))=sqrt(12.141564 xx 4.46450 xx 6.76048 xx 0.92067) = 18.371`

`S_("parallelogram ABDC") = 2*S_(triangle ABC) = 2*18.371 = 36.712`

Other way to find the area of the kind of triangle involved in this question is described in:
when the triangle is embedded in three-dimensional space