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also see Defining Geometric Figures


  A polygon (plane figure) with 4 angles
and 4 sides.

Sides: a, b, c, d
Angles: A, B, C, D
Around the quadrilateral are a, A, b, B, c, C, d, D, and back to a, in that order

Altitudes: ha , etc.
Diagonals: p = BD, q = AC, intersect at O
Angle between diagonals: theta

Perimeter: P
Semiperimeter: s
Area: K

Radius of circumscribed circle: R
Radius of inscribed circle: r


To read about quadrilaterals, visit The Geometry Center. 



P = a + b + c + d.
s = P/2 = (a+b+c+d)/2
A + B + C + D = 2 Pi radians = 360o
K = pq sin(theta)/2
K = (b2+d2-a2-c2)tan(theta)/4
K = sqrt[4p2q2-(b2+d2-a2-c2)2]/4
K = sqrt[(s-a)(s-b)(s-c)(s-d)-abcd
     (Bretschneider's Formula)



A quadrilateral with four right angles and all four sides of equal length.

a = b = c = d
A = B = C = D = Pi/2 radians = 90o
theta = Pi/2 radians = 90o

    ha = a
p = q = a sqrt(2)
P = 4a
s = 2a
K = a2
R = a sqrt(2)/2
r = a/2

JavaSketchpad exploration: Square


A quadrilateral with adjacent sides perpendicular
(all four angles are therefore right angles).

a = c, b = d.
A = B = C = D = Pi/2 radians = 90o

ha = b
hb = a
p = q = sqrt(a2+b2)
theta = 2 arctan(a/b)
P = 2(a+b)
s = a + b
K = ab
R = p/2 = sqrt(a2+b2)/2
r = minimum(a,b)/2

JavaSketchpad exploration:

A quadrilateral with opposite sides parallel.

a = c, b = d
A = C, B = D


    A + B = Pi radians = 180o
ha = b sin(A) = b cos(B-Pi/2)
hb = a sin(A) = a cos(B-Pi/2)
p = sqrt[a2+b2-2ab cos(A)]
q = sqrt[a2+b2-2ab cos(B)]
p2+q2 = 2(a2+b2)
theta = arccos([a2-b2]/pq)
P = 2*(a+b)
s = a + b
K = ab sin(A) = ab sin(B) = bhb
   = pq sin(theta)/2

JavaSketchpad exploration: Parallelogram


A parallelogram with all sides equal.

a = b = c = d
A = C, B = D
theta = Pi/2 radians = 90o

    A + B = Pi radians = 180o
ha = a sin(A) = a cos(B-Pi/2)
ha = hb
p = a sqrt[2-2 cos(A)]
q = a sqrt[2-2 cos(B)]
p2+q2 = 4a2
P = 4a
s = 2a
K = a2sin(A) = a2sin(B)
     = aha = pq/2

JavaSketchpad exploration: Rhombus

  Trapezoid (American)
  Trapezium (British)

a parallel to c, m = (a+c)/2
A + B = C + D = Pi radians = 180o
        P = a + b + c + d

    K = ham = ha(a+c)/2

  If a = c, the trapezoid is actually a parallelogram, so b = d, and the height and area cannot be determined from a, b, c, and d alone. If a and c are not equal, then

        ha2 = (a+b-c+d)(-a+b+c+d)(a-b-c+d)(a+b-c-d)/[4(a-c)2].

  If ha2 < 0, no trapezoid having those side lengths exists.

*From The Words of Mathematics by Steven Schwartzman (1994, Mathematical Association of America):

trapezoid (noun); trapezoidal (adjective); trapezium, plural trapezia (noun): The Greek word trapeza "table" was composed of tetra "four" and the Indo-European root ped- "foot." A Greek table must have had four feet (= legs). The suffix -oid (q.v.) means "looking like," so that a trapezoid is a figure that looks like a table (at least in somebody's imagination). Some Americans define a trapezoid as a quadrilateral with at least one pair of parallel sides. Under that definition, a parallelogram is a special kind of trapezoid. For other Americans, however, a trapezoid is a quadrilateral with one and only one pair of parallel sides, in which case a parallelogram is not a trapezoid. The situation is further confused by the fact that in Europe a trapezoid is defined as a quadrilateral with no sides equal. Even more confusing is the existence of the similar word trapezium, which in American usage means "a quadrilateral with no sides equal," but which in European usage is a synonym of what Americans call a trapezoid. Apparently to cut down on the confusion, trapezium is not used in American textbooks. The trapeze used in a circus is also related, since a trapeze has or must once have had four "sides": two ropes, the bar at the bottom, and a support bar at the top.


JavaSketchpad exploration: Trapezoid


A quadrilateral with two pairs of distinct
adjacent sides equal in length.

a = b, c = d
theta = Pi/2 radians = 90o

    OB = OD = p/2, OA = h, OC = q - h
h = sqrt(a2-p2/4)
q = sqrt(a2-p2/4) + sqrt(c2-p2/4)
P = 2(a+c)
K = pq/2

JavaSketchpad exploration: Kite

  Cyclic Quadrilateral

A quadrilateral all of whose vertices lie on a circle.

Points A, B, C, and D lie on a circle of radius R.
A + C = B + D = Pi radians = 180o

    K = sqrt[(s-a)(s-b)(s-c)(s-d)]
     (Brahmagupta's Formula)
K = sqrt[(ab+cd)(ac+bd)(ad+bc)]/4*R
p = sqrt[(ac+bd)(ad+bc)/(ab+cd)]
q = sqrt[(ab+cd)(ac+bd)/(ad+bc)]
R = sqrt[(ab+cd)(ac+bd)(ad+bc) /
theta = arcsin[2K/(ac+bd)]


A quadrilateral within which a circle can be
inscribed, tangent to all four sides.

Points A, B, C, and D lie on a circle of radius R.
Sides a, b, c, and d are tangent to a circle of radius r.
m = distance between the centers of the two circles.

A + C = B + D = Pi radians = 180o
a + c = b + d

K = sqrt[abcd]

r = sqrt[abcd]/s
R = sqrt[(ab+cd)(ac+bd)(ad+bc)/abcd]/4
1/(R+m)2 + 1/(R-m)2 = 1/r2

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