| Parallelogram | |
| Type: | quadrilateral, trapezium |
| Edges: | 4 |
| Symmetry: | C2, [2]+, |
| Area: | b × h (base × height); ab sin θ (product of adjacent sides and sine of the vertex angle determined by them) |
| Properties: | convex |
In Euclidean geometry, a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. The congruence of opposite sides and opposite angles is a direct consequence of the Euclidean parallel postulate and neither condition can be proven without appealing to the Euclidean parallel postulate or one of its equivalent formulations.
By comparison, a quadrilateral with at least one pair of parallel sides is a trapezoid in American English or a trapezium in British English.
The three-dimensional counterpart of a parallelogram is a parallelepiped.
The word comes from the Greek παραλληλό-γραμμον, parallēló-grammon, which means a shape "of parallel lines".
A simple (non-self-intersecting) quadrilateral is a parallelogram if and only if any one of the following statements is true:[2] [3]
Thus all parallelograms have all the properties listed above, and conversely, if just one of these statements is true in a simple quadrilateral, then it is a parallelogram.
All of the 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:
K=bh.
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
Krect=(B+A) x H
and the area of a single triangle is
Ktri=
| A | |
| 2 |
x H.
Therefore, the area of the parallelogram is
K=Krect-2 x Ktri=((B+A) x H)-(A x H)=B x H.
Another area formula, for two sides B and C and angle θ, is
K=B ⋅ C ⋅ \sin\theta.
The area of a parallelogram with sides B and C (B ≠ C) and angle
\gamma
K=
| |\tan\gamma| | |
| 2 |
⋅ \left|B2-C2\right|.
When the parallelogram is specified from the lengths B and C of two adjacent sides together with the length D1 of either diagonal, then the area can be found from Heron's formula. Specifically it is
| K=2\sqrt{S(S-B)(S-C)(S-D | ||||
|
\sqrt{(B+C+D)(-B+C+D)(B-C+D)(B+C-D)},
where
S=(B+C+D1)/2
Let vectors
a,b\in\R2
V=\begin{bmatrix}a1&a2\ b1&b2\end{bmatrix}\in\R2
|\det(V)|=|a1b2-a2b1|
Let vectors
a,b\in\Rn
V=\begin{bmatrix}a1&a2&...&an\ b1&b2&...&bn\end{bmatrix}\in\R2
\sqrt{\det(VVT)}
Let points
a,b,c\in\R2
K=\left|\begin{matrix} a1&a2&1\\ b1&b2&1\\ c1&c2&1 \end{matrix}\right|.
To prove that the diagonals of a parallelogram bisect each other, we will use congruent triangles:
\angleABE\cong\angleCDE
\angleBAE\cong\angleDCE
(since these are angles that a transversal makes with parallel lines AB and DC).
Also, side AB is equal in length to side DC, since opposite sides of a parallelogram are equal in length.
Therefore, triangles ABE and CDE are congruent (ASA postulate, two corresponding angles and the included side).
Therefore,
AE=CE
BE=DE.
Since the diagonals AC and BD divide each other into segments of equal length, the diagonals bisect each other.
Separately, since the diagonals AC and BD bisect each other at point E, point E is the midpoint of each diagonal.
Parallelograms can tile the plane by translation. If edges are equal, or angles are right, the symmetry of the lattice is higher. These represent the four Bravais lattices in 2 dimensions.
| Form | Square | Rectangle | Rhombus | Rhomboid | |
|---|---|---|---|---|---|
| System | Square (tetragonal) | Rectangular (orthorhombic) | Centered rectangular (orthorhombic) | Oblique (monoclinic) | |
| Constraints | α=90°, a=b | α=90° | a=b | None | |
| Symmetry | p4m, [4,4], order 8n | pmm, [∞,2,∞], order 4n | p1, [∞<sup>+</sup>,2,∞<sup>+</sup>], order 2n | ||
| Form | |||||
An automedian triangle is one whose medians are in the same proportions as its sides (though in a different order). If ABC is an automedian triangle in which vertex A stands opposite the side a, G is the centroid (where the three medians of ABC intersect), and AL is one of the extended medians of ABC with L lying on the circumcircle of ABC, then BGCL is a parallelogram.
See main article: Varignon's theorem.
Varignon's theorem holds that the midpoints of the sides of an arbitrary quadrilateral are the vertices of a parallelogram, called its Varignon parallelogram. If the quadrilateral is convex or concave (that is, not self-intersecting), then the area of the Varignon parallelogram is half the area of the quadrilateral.
Proof without words (see figure):
For an ellipse, two diameters are said to be conjugate if and only if the tangent line to the ellipse at an endpoint of one diameter is parallel to the other diameter. Each pair of conjugate diameters of an ellipse has a corresponding tangent parallelogram, sometimes called a bounding parallelogram, formed by the tangent lines to the ellipse at the four endpoints of the conjugate diameters. All tangent parallelograms for a given ellipse have the same area.
It is possible to reconstruct an ellipse from any pair of conjugate diameters, or from any tangent parallelogram.
A parallelepiped is a three-dimensional figure whose six faces are parallelograms.