Editing Parallelogram (section)

==Area formula{{anchor|Area}}==
[[File:ParallelogramArea.svg|thumb|180px|alt=A diagram showing how a parallelogram can be re-arranged into the shape of a rectangle|A parallelogram can be rearranged into a rectangle with the same area.]]
[[File:Parallelogram area animated.gif|thumb|180px|Animation for the area formula <math>K = bh</math>.]]
All of the [[Quadrilateral#Area of a convex quadrilateral|area formulas for general convex quadrilaterals]] apply to parallelograms. Further formulas are specific to parallelograms:

A parallelogram with base ''b'' and height ''h'' can be divided into a [[trapezoid]] and a [[right triangle]], and rearranged into a [[rectangle]], as shown in the figure to the left. This means that the [[area]] of a parallelogram is the same as that of a rectangle with the same base and height:

:<math>K = bh.</math>

[[File:Parallelogram area.svg|thumb|The area of the parallelogram is the area of the blue region, which is the interior of the parallelogram]]
The base × height area formula can also be derived using the figure to the right. The area ''K'' of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is

:<math>K_\text{rect} = (B+A) \times H\,</math>

and the area of a single triangle is

:<math>K_\text{tri} = \frac{A}{2} \times H. \,</math>

Therefore, the area of the parallelogram is

:<math>K = K_\text{rect} - 2 \times K_\text{tri} = ( (B+A) \times H) - ( A \times H) = B \times H.</math>

Another area formula, for two sides ''B'' and ''C'' and angle θ, is

:<math>K = B \cdot C \cdot \sin \theta.\,</math>

Provided that the parallelogram is a rhombus, the area can be expressed using sides ''B'' and ''C'' and angle <math>\gamma</math> at the intersection of the diagonals:<ref>Mitchell, Douglas W., "The area of a quadrilateral", ''Mathematical Gazette'', July 2009.</ref>

:<math>K = \frac{|\tan \gamma|}{2} \cdot \left| B^2 - C^2 \right|.</math>


When the parallelogram is specified from the lengths ''B'' and ''C'' of two adjacent sides together with the length ''D''<sub>1</sub> of either diagonal, then the area can be found from [[Heron's formula]]. Specifically it is

:<math>K=2\sqrt{S(S-B)(S-C)(S-D_1)}=\frac{1}{2}\sqrt{(B+C+D_1)(-B+C+D_1)(B-C+D_1)(B+C-D_1)},</math>

where <math>S=(B+C+D_1)/2</math> and the leading factor 2 comes from the fact that the chosen diagonal divides the parallelogram into ''two'' congruent triangles.

=== From vertex coordinates ===
Let vectors <math>\mathbf{a},\mathbf{b}\in\R^2</math> and let <math>V = \begin{bmatrix} a_1 & a_2 \\ b_1 & b_2 \end{bmatrix} \in\R^{2 \times 2}</math> denote the matrix with elements of '''a''' and '''b'''. Then the area of the parallelogram generated by '''a''' and '''b''' is equal to <math>|\det(V)| = |a_1b_2 - a_2b_1|\,</math>.

Let vectors <math>\mathbf{a},\mathbf{b}\in\R^n</math> and let <math>V = \begin{bmatrix} a_1 & a_2 & \dots & a_n \\ b_1 & b_2 & \dots & b_n \end{bmatrix} \in\R^{2 \times n}</math>. Then the area of the parallelogram generated by '''a''' and '''b''' is equal to <math>\sqrt{\det(V V^\mathrm{T})}</math>.

Let points <math>a,b,c\in\R^2</math>. Then the [[signed area]] of the parallelogram with vertices at ''a'', ''b'' and ''c'' is equivalent to the determinant of a matrix built using ''a'', ''b'' and ''c'' as rows with the last column padded using ones as follows:
:<math>K = \left| \begin{matrix}
        a_1 & a_2 & 1 \\
        b_1 & b_2 & 1 \\
        c_1 & c_2 & 1
 \end{matrix} \right|. </math>