How do you find the area of a parallelogram with vertices?

2 Answers
Feb 14, 2016

For parallelogram #ABCD# the area is
#S = |(x_B-x_A)*(y_D-y_A)-(y_B-y_A)*(x_D-x_A)|#

Explanation:

Let's assume that our parallelogram #ABCD# is defined by the coordinates of its four vertices - #[x_A,y_A]#, #[x_B,y_B]#, #[x_C,y_C]#, #[x_D,y_D]#.

To determine the area of our parallelogram, we need the length of its base #|AB|# and the altitude #|DH|# from vertex #D# to point #H# on side #AB# (that is, #DH_|_AB#).

First of all, to simplify the task, let's move it to a position when its vertex #A# coincides with the origin of coordinates. The area will be the same, but calculations will be easier.
So, we will perform the following transformation of coordinates:
#U=x-x_A#
#V=y-y_A#

Then the (#U,V#) coordinates of all vertices will be:
#A[U_A=0,V_B=0]#
#B[U_B=x_B-x_A,V_B=y_B-y_A]#
#C[U_C=x_C-x_A,V_C=y_C-y_A]#
#D[U_D=x_D-x_A,V_D=y_D-y_A]#

Our parallelogram now is defined by two vectors:
#p=(U_B,V_B)# and #q=(U_D,V_D)#

Determine the length of base #AB# as the length of vector #p#:
#|AB|=sqrt(U_B^2+V_B^2)#

The length of altitude #|DH|# can be expressed as #|AD|*sin(/_BAD)#.
The length #AD# is the length of vector #q#:
#|AD|=sqrt(U_D^2+V_D^2)#
Angle #/_BAD# can be determined by using two expressions for the scalar (dot) product of vectors #p# and #q#:
#(p*q)=U_B*U_D+V_B*V_D=|p|*|q|*cos(/_BAD)#
from which
#cos^2(/_BAD)=(U_B*U_D+V_B*V_D)^2/[(U_B^2+V_B^2)*(U_D^2+V_D^2)]#
#sin^2(/_BAD)=1-cos^2(/_BAD)=#
#=1-(U_B*U_D+V_B*V_D)^2/[(U_B^2+V_B^2)*(U_D^2+V_D^2)]=#
# =(U_B*V_D-V_B*U_D)^2 / [(U_B^2+V_B^2)*(U_D^2+V_D^2)]#

Now we know all components to calculate the area:
Base #|AB|=sqrt(U_B^2+V_B^2)#:
Altitude #|DH|=sqrt(U_D^2+V_D^2)*|U_A*V_D-V_A*U_D| / sqrt[(U_B^2+V_B^2)*(U_D^2+V_D^2)]#

The area is their product:
#S = |AB|*|DH|=|U_B*V_D-V_B*U_D|#

In terms of original coordinates, it looks like this:
#S = |(x_B-x_A)*(y_D-y_A)-(y_B-y_A)*(x_D-x_A)|#

Mar 4, 2016

another discussion

Explanation:

Geometric proof
Considering the figure
enter image source here

we can easily establish the formula for calculation of the area of a parallelogram ABCD, when any three vertices (say A,B,D) are known.

Since diagonal BD bisects the parallelogram into two congruent triangle.
The area of the parallelogram ABCD
= 2 area of triangle ABD
=2[ area of trapezium BAPQ +area of trap BQRD - area of trap DAPR]
=2[#1/2(AP+BQ)PQ+1/2(BQ+DR)QR-1/2(AP+DR)PR]#
= #(Y_A+Y_B)(X_B-X_A )+(Y_B+Y_D)(X_D-X_B)-(Y_A+Y_D)(X_D-X_A)#
=#Y_AX_B+cancel (Y_BX_B)-cancel(Y_AX_A)-Y_BX_A +Y_BX_D+cancel(Y_DX_D)-cancel(Y_BX_B)-Y_AX_D-cancel(Y_DX_D)+cancel(Y_AX_A)+Y_DX_A#

=#Y_A(X_B_X_D)+Y_B(X_D-XA)+Y_D(X_A-X_B)#

This formula will give the area of the parallelogram .
Proof considering vector
It can also be established considering #vec(AB)# and# vec(AD)#
Now
Position vector of point A w.r,t the origin O, #vec(OA)= X_Ahati +Y_Ahatj#
Position vector of point B w.r,t the origin O, #vec(OB)= X_Bhati +Y_Bhatj#
Position vector of point D w.r,t the origin O, #vec(OD)= X_Dhati +Y_Dhatj#

Now
Area of the Parallelogram ABCD
#= Base (AD)*Height (BE)=AD*h#
#=AD*ABsintheta=|vec(AD)Xvec(AB)|#

Again
#vec(AD)=vec(OD)-vec(OA)=(X_D-X_A)hati+(Y_D-Y_A)hatj#
#vec(AB)=vec(OB)-vec(OA)=(X_B-X_A)hati+(Y_B-Y_A)hatj#
#vec(AD)#X#vec(AB)=[(X_D-X_A)(Y_B-Y_A)-(X_B-X_A)(Y_D-Y_A)]hatk#
Area = #|vec(AD)#X#vec(AB)|#
=# |Y_BX_D-Y_BX_A-Y_AX_D+cancel(Y_AX_A)-Y_DX_B +Y_DX_A+Y_AX_B-cancel(Y_AX_A)|#
=# |Y_BX_D-Y_BX_A-Y_AX_D-Y_DX_B +Y_DX_A+Y_AX_B|#
=# |Y_BX_D-Y_BX_A-Y_AX_D-Y_DX_B +Y_DX_A+Y_AX_B|#
=#|Y_A(X_B_X_D)+Y_B(X_D-XA)+Y_D(X_A-X_B)|#
Thus we have the same formula