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)Xvec(AB)=[(X_D-X_A)(Y_B-Y_A)-(X_B-X_A)(Y_D-Y_A)]hatk
Area = |vec(AD)Xvec(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