Find the area of a parallelogram with vertices A(1,5,0), B(6,10,−3), C(−4,5,−2), and D(1,10,−5)? Show steps.

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Rui D. Share
Jan 6, 2016

Answer:

#S_("parallelogram") = 15*sqrt(6)~=36.742#

Explanation:

Note that it doesn't matter whether the parallelogram is formed by the vertices A, B, C, D in this order or by the vertices A, B, D. C in this order, any diagonal of the parallelogram divides it in two triangles with equal areas. Then #S_("parallelogram" =2*S_(triangle)#. In this explanation the #triangle ABC# is selected (but any other possible triangle formed with the points A, B. C or D, would do).
Repeating the coordinates of the points of #triangle ABC#:
A(1,5,0)
B(6,10,-3)
C(-4,5,-2)

Obtaining sides of the triangle ABC:
#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)#

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") = 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

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