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

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

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Answer 1 · 2016-02-14T15:05:36

For parallelogram $A B C D$ $ABCD$ the area is
$S = | (x_{B} - x_{A}) \cdot (y_{D} - y_{A}) - (y_{B} - y_{A}) \cdot (x_{D} - x_{A}) |$ $S = |(x_B-x_A)*(y_D-y_A)-(y_B-y_A)*(x_D-x_A)|$

Explanation:

Let's assume that our parallelogram $A B C D$ $ABCD$ is defined by the coordinates of its four vertices - $[x_{A}, y_{A}]$ $[x_A,y_A]$ , $[x_{B}, y_{B}]$ $[x_B,y_B]$ , $[x_{C}, y_{C}]$ $[x_C,y_C]$ , $[x_{D}, y_{D}]$ $[x_D,y_D]$ .

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

First of all, to simplify the task, let's move it to a position when its vertex $A$ $A$ 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:
$U = x - x_{A}$ $U=x-x_A$
$V = y - y_{A}$ $V=y-y_A$

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

Our parallelogram now is defined by two vectors:
$p = (U_{B}, V_{B})$ $p=(U_B,V_B)$ and $q = (U_{D}, V_{D})$ $q=(U_D,V_D)$

Determine the length of base $A B$ $AB$ as the length of vector $p$ $p$ :
$| A B | = \sqrt{U_{B}^{2} + V_{B}^{2}}$ $|AB|=sqrt(U_B^2+V_B^2)$

The length of altitude $| D H |$ $|DH|$ can be expressed as $| A D | \cdot sin (∠ B A D)$ $|AD|*sin(/_BAD)$ .
The length $A D$ $AD$ is the length of vector $q$ $q$ :
$| A D | = \sqrt{U_{D}^{2} + V_{D}^{2}}$ $|AD|=sqrt(U_D^2+V_D^2)$
Angle $∠ B A D$ $/_BAD$ can be determined by using two expressions for the scalar (dot) product of vectors $p$ $p$ and $q$ $q$ :
$(p \cdot q) = U_{B} \cdot U_{D} + V_{B} \cdot V_{D} = | p | \cdot | q | \cdot cos (∠ B A D)$ $(p*q)=U_B*U_D+V_B*V_D=|p|*|q|*cos(/_BAD)$
from which
${cos}^{2} (∠ B A D) = \frac{{(U_{B} \cdot U_{D} + V_{B} \cdot V_{D})}_{B}^{2}}{(U_{B}^{2} + V_{B}^{2}) \cdot (U_{D}^{2} + V_{D}^{2})}$ $cos^2(/_BAD)=(U_B*U_D+V_B*V_D)^2/[(U_B^2+V_B^2)*(U_D^2+V_D^2)]$
${sin}^{2} (∠ B A D) = 1 - {cos}^{2} (∠ B A D) =$ $sin^2(/_BAD)=1-cos^2(/_BAD)=$
$= 1 - \frac{{(U_{B} \cdot U_{D} + V_{B} \cdot V_{D})}_{B}^{2}}{(U_{B}^{2} + V_{B}^{2}) \cdot (U_{D}^{2} + V_{D}^{2})} =$ $=1-(U_B*U_D+V_B*V_D)^2/[(U_B^2+V_B^2)*(U_D^2+V_D^2)]=$
$= \frac{{(U_{B} \cdot V_{D} - V_{B} \cdot U_{D})}_{B}^{2}}{(U_{B}^{2} + V_{B}^{2}) \cdot (U_{D}^{2} + V_{D}^{2})}$ $=(U_B*V_D-V_B*U_D)^2 / [(U_B^2+V_B^2)*(U_D^2+V_D^2)]$

Now we know all components to calculate the area:
Base $| A B | = \sqrt{U_{B}^{2} + V_{B}^{2}}$ $|AB|=sqrt(U_B^2+V_B^2)$ :
Altitude $| D H | = \sqrt{U_{D}^{2} + V_{D}^{2}} \cdot \frac{| U_{A} \cdot V_{D} - V_{A} \cdot U_{D} |}{\sqrt{(U_{B}^{2} + V_{B}^{2}) \cdot (U_{D}^{2} + V_{D}^{2})}}$ $|DH|=sqrt(U_D^2+V_D^2)*|U_A*V_D-V_A*U_D| / sqrt[(U_B^2+V_B^2)*(U_D^2+V_D^2)]$

The area is their product:
$S = | A B | \cdot | D H | = | U_{B} \cdot V_{D} - V_{B} \cdot U_{D} |$ $S = |AB|*|DH|=|U_B*V_D-V_B*U_D|$

In terms of original coordinates, it looks like this:
$S = | (x_{B} - x_{A}) \cdot (y_{D} - y_{A}) - (y_{B} - y_{A}) \cdot (x_{D} - x_{A}) |$ $S = |(x_B-x_A)*(y_D-y_A)-(y_B-y_A)*(x_D-x_A)|$

Answer 2 · 2016-03-04T14:05:19

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[ $\frac{1}{2} (A P + B Q) P Q + \frac{1}{2} (B Q + D R) Q R - \frac{1}{2} (A P + D R) P R]$ $1/2(AP+BQ)PQ+1/2(BQ+DR)QR-1/2(AP+DR)PR]$
= $(Y_{A} + Y_{B}) (X_{B} - X_{A}) + (Y_{B} + Y_{D}) (X_{D} - X_{B}) - (Y_{A} + Y_{D}) (X_{D} - X_{A})$ $(Y_A+Y_B)(X_B-X_A )+(Y_B+Y_D)(X_D-X_B)-(Y_A+Y_D)(X_D-X_A)$
= $Y_AX_B+cancel (Y_BX_B)-cancel(Y_AX_A)-Y_BX_A +Y_BX_D+cancel(Y_DX_D)-cancel(Y_BX_B)-Y_AX_D-cancel(Y_DX_D)+cancel(Y_AX_A)+Y_DX_A$

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

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

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

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