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)?

1 Answer
Jan 1, 2016

#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