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)AB=√(6−1)2+(10−5)2+(−3−0)2=√25+25+9=√59
AC=sqrt((-4-1)^2+(5-5)^2+(-2-0)^2)=sqrt(25+0+4)=sqrt(29)AC=√(−4−1)2+(5−5)2+(−2−0)2=√25+0+4=√29
BC=sqrt((-4-6)^2+(5-10)^2+(-2+3)^2)=sqrt(100+25+1)=sqrt(126)BC=√(−4−6)2+(5−10)2+(−2+3)2=√100+25+1=√126
CD=sqrt((1+4)^2+(10-5)^2+(-5+2)^2)=sqrt(25+25+9)=sqrt(59)CD=√(1+4)2+(10−5)2+(−5+2)2=√25+25+9=√59
DA=sqrt((1-1)^2+(5-10)^2+(0+5)^2)=sqrt(0+25+25)=sqrt(50)DA=√(1−1)2+(5−10)2+(0+5)2=√0+25+25=√50
Just a minute! AB=CDAB=CD but BC!=DA!BC≠DA! Let's calculate also BD:
BD=sqrt((1-6)^2+(10-10)^2+(-5+3)^2)=sqrt(25+0+4)=sqrt(29)BD=√(1−6)2+(10−10)2+(−5+3)2=√25+0+4=√29
Since AB=CDAB=CD and AC=BDAC=BD the parallelogram should be called ABDC , composed of the triangles ABC and ABD, and, because these triangles are congruent, S_(triangleABC) = S_(triangleBDC)S△ABC=S△BDC and S_("parallelogram" ABDC) = 2*S_(triangleABC)SparallelogramABDC=2⋅S△ABC
Using Heron's Formula
Triangle ABC (a=AB, b=AC and c=BCa=AB,b=ACandc=BC)
s=(a+b+c)/2=(sqrt(59)+sqrt(29)+sqrt(126))/2~=12.14564s=a+b+c2=√59+√29+√1262≅12.14564
(s-a)=(-sqrt(59)+sqrt(29)+sqrt(126))/2~=4.46450(s−a)=−√59+√29+√1262≅4.46450
(s-b)=(sqrt(59)-sqrt(29)+sqrt(126))/2~=6.76048(s−b)=√59−√29+√1262≅6.76048
(s-c)=(sqrt(59)+sqrt(29)-sqrt(126))/2~=0.92067(s−c)=√59+√29−√1262≅0.92067
S_(triangleABC)=sqrt(s(s-a)(s-b)(s-c))=sqrt(12.141564 xx 4.46450 xx 6.76048 xx 0.92067) = 18.371S△ABC=√s(s−a)(s−b)(s−c)=√12.141564×4.46450×6.76048×0.92067=18.371
S_("parallelogram ABDC") = 2*S_(triangle ABC) = 2*18.371 = 36.712Sparallelogram ABDC=2⋅S△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
