In geometry a quadrilateral is a four-sided polygon, having four edges (sides) and four corners (vertices).

[2] All non-self-crossing quadrilaterals tile the plane, by repeated rotation around the midpoints of their edges.

[10] The two diagonals of a convex quadrilateral are the line segments that connect opposite vertices.

The two bimedians of a convex quadrilateral are the line segments that connect the midpoints of opposite sides.

The four maltitudes of a convex quadrilateral are the perpendiculars to a side—through the midpoint of the opposite side.

[13] There are various general formulas for the area K of a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD and d = DA.

The area can be expressed in trigonometric terms as[14] where the lengths of the diagonals are p and q and the angle between them is θ.

[15] In the case of an orthodiagonal quadrilateral (e.g. rhombus, square, and kite), this formula reduces to

In a parallelogram, where both pairs of opposite sides and angles are equal, this formula reduces to

Another area formula including the sides a, b, c, d is[16] where x is the distance between the midpoints of the diagonals, and φ is the angle between the bimedians.

The last trigonometric area formula including the sides a, b, c, d and the angle α (between a and b) is:[19] which can also be used for the area of a concave quadrilateral (having the concave part opposite to angle α), by just changing the first sign + to -.

[26] The list applies to the most general cases, and excludes named subsets.

The German mathematician Carl Anton Bretschneider derived in 1842 the following generalization of Ptolemy's theorem, regarding the product of the diagonals in a convex quadrilateral[28] This relation can be considered to be a law of cosines for a quadrilateral.

If X and Y are the feet of the normals from B and D to the diagonal AC = p in a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD, d = DA, then[29]: p.14 In a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD, d = DA, and where the diagonals intersect at E, where e = AE, f = BE, g = CE, and h = DE.

[30] The shape and size of a convex quadrilateral are fully determined by the lengths of its sides in sequence and of one diagonal between two specified vertices.

[31] The bimedians of a quadrilateral are the line segments connecting the midpoints of the opposite sides.

The lengths of the bimedians can also be expressed in terms of two opposite sides and the distance x between the midpoints of the diagonals.

The area of any quadrilateral also satisfies the inequality[38] Denoting the perimeter as L, we have[38]: p.114 with equality only in the case of a square.

Let a, b, c, d be the lengths of the sides of a convex quadrilateral ABCD with the area K and diagonals AC = p, BD = q.

The dual theorem states that of all quadrilaterals with a given area, the square has the shortest perimeter.

[38]: p.119  This is a direct consequence of the fact that the area of a convex quadrilateral satisfies where θ is the angle between the diagonals p and q.

The "vertex centroid" comes from considering the quadrilateral as being empty but having equal masses at its vertices.

Let Ga, Gb, Gc, Gd be the centroids of triangles BCD, ACD, ABD, ABC respectively.

[47] In a general convex quadrilateral ABCD, there are no natural analogies to the circumcenter and orthocenter of a triangle.

Let Oa, Ob, Oc, Od be the circumcenters of triangles BCD, ACD, ABD, ABC respectively; and denote by Ha, Hb, Hc, Hd the orthocenters in the same triangles.

In a convex quadrilateral, the quasiorthocenter H, the "area centroid" G, and the quasicircumcenter O are collinear in this order, and HG = 2GO.

[47] There can also be defined a quasinine-point center E as the intersection of the lines EaEc and EbEd, where Ea, Eb, Ec, Ed are the nine-point centers of triangles BCD, ACD, ABD, ABC respectively.

[49] For a convex quadrilateral ABCD in which E is the point of intersection of the diagonals and F is the point of intersection of the extensions of sides BC and AD, let ω be a circle through E and F which meets CB internally at M and DA internally at N. Let CA meet ω again at L and let DB meet ω again at K. Then there holds: the straight lines NK and ML intersect at point P that is located on the side AB; the straight lines NL and KM intersect at point Q that is located on the side CD.

Note that "trapezoid" here is referring to the North American definition (the British equivalent is a trapezium).

Formulas to compute its dihedral angles from the edge lengths and the angle between two adjacent edges were derived for work on the properties of molecules such as cyclobutane that contain a "puckered" ring of four atoms.