Isosceles trapezoid

 

Rectangles and squares are usually considered to be special cases of isosceles trapezoids though some sources would exclude them.^[3]

Another special case is a 3-equal side trapezoid, sometimes known as a trilateral trapezoid^[4] or a trisosceles trapezoid. They can also be seen dissected from regular polygons of 5 sides or more as a truncation of 4 sequential vertices.

Any non-self-crossing quadrilateral with exactly one axis of symmetry must be either an isosceles trapezoid or a kite.^[5] However, if crossings are allowed, the set of symmetric quadrilaterals must be expanded to include also the crossed isosceles trapezoids, crossed quadrilaterals in which the crossed sides are of equal length and the other sides are parallel, and the antiparallelograms, crossed quadrilaterals in which opposite sides have equal length.

Every antiparallelogram has an isosceles trapezoid as its convex hull, and may be formed from the diagonals and non-parallel sides (or either pair of opposite sides in the case of a rectangle) of an isosceles trapezoid.^[6]

Convex isosceles trapezoid	Crossed isosceles trapezoid	antiparallelogram

If a quadrilateral is known to be a trapezoid, it is not sufficient just to check that the legs have the same length in order to know that it is an isosceles trapezoid, since a rhombus is a special case of a trapezoid with legs of equal length, but is not an isosceles trapezoid as it lacks a line of symmetry through the midpoints of opposite sides.

Any one of the following properties distinguishes an isosceles trapezoid from other trapezoids:

• The diagonals have the same length.
• The base angles have the same measure.
• The segment that joins the midpoints of the parallel sides is perpendicular to them.
• Opposite angles are supplementary, which in turn implies that isosceles trapezoids are cyclic quadrilaterals.
• The diagonals divide each other into segments with lengths that are pairwise equal; in terms of the picture below, AE = DE, BE = CE (and AE ≠ CE if one wishes to exclude rectangles).

In an isosceles trapezoid, the base angles have the same measure pairwise. In the picture below, angles ∠ABC and ∠DCB are obtuse angles of the same measure, while angles ∠BAD and ∠CDA are acute angles, also of the same measure.

Since the lines AD and BC are parallel, angles adjacent to opposite bases are supplementary, that is, angles ∠ABC + ∠BAD = 180°.

 

The diagonals of an isosceles trapezoid have the same length; that is, every isosceles trapezoid is an equidiagonal quadrilateral. Moreover, the diagonals divide each other in the same proportions. As pictured, the diagonals AC and BD have the same length (AC = BD) and divide each other into segments of the same length (AE = DE and BE = CE).

The ratio in which each diagonal is divided is equal to the ratio of the lengths of the parallel sides that they intersect, that is,

${\displaystyle {\frac {AE}{EC}}={\frac {DE}{EB}}={\frac {AD}{BC}}.}$ 

The length of each diagonal is, according to Ptolemy's theorem, given by

${\displaystyle p={\sqrt {ab+c^{2}}}}$ 

where a and b are the lengths of the parallel sides AD and BC, and c is the length of each leg AB and CD.

The height is, according to the Pythagorean theorem, given by

${\displaystyle h={\sqrt {p^{2}-\left({\frac {a+b}{2}}\right)^{2}}}={\tfrac {1}{2}}{\sqrt {4c^{2}-(a-b)^{2}}}.}$ 

The distance from point E to base AD is given by

${\displaystyle d={\frac {ah}{a+b}}}$ 

where a and b are the lengths of the parallel sides AD and BC, and h is the height of the trapezoid.

The area of an isosceles (or any) trapezoid is equal to the average of the lengths of the base and top (the parallel sides) times the height. In the adjacent diagram, if we write AD = a, and BC = b, and the height h is the length of a line segment between AD and BC that is perpendicular to them, then the area K is

${\displaystyle K={\tfrac {1}{2}}\left(a+b\right)h.}$ 

If instead of the height of the trapezoid, the common length of the legs AB =CD = c is known, then the area can be computed using Brahmagupta's formula for the area of a cyclic quadrilateral, which with two sides equal simplifies to

${\displaystyle K=(s-c){\sqrt {(s-a)(s-b)}},}$ 

where ${\displaystyle s={\tfrac {1}{2}}(a+b+2c)}$  is the semi-perimeter of the trapezoid. This formula is analogous to Heron's formula to compute the area of a triangle. The previous formula for area can also be written as

${\displaystyle K={\frac {a+b}{4}}{\sqrt {(a-b+2c)(b-a+2c)}}.}$ 

The radius in the circumscribed circle is given by^[7]

${\displaystyle R=c{\sqrt {\frac {ab+c^{2}}{4c^{2}-(a-b)^{2}}}}.}$ 

In a rectangle where a = b this is simplified to ${\displaystyle R={\tfrac {1}{2}}{\sqrt {a^{2}+c^{2}}}}$ .

• Isosceles tangential trapezoid

• Some engineering formulas involving isosceles trapezoids