Quadrilateral



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: h_{a} , 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.

General 



P = a + b + c + d.
s = P/2 = (a+b+c+d)/2
A + B + C + D = 2 Pi radians = 360^{o}
K = pq sin(theta)/2
K = (b^{2}+d^{2}a^{2}c^{2})tan(theta)/4
K = sqrt[4p^{2}q^{2}(b^{2}+d^{2}a^{2}c^{2})^{2}]/4
K = sqrt[(sa)(sb)(sc)(sd)abcd
cos^{2}([A+C]/2)]
(Bretschneider's Formula) 
Square 



A quadrilateral with four right angles and all four sides of equal
length.
a = b = c = d
A = B = C = D = Pi/2 radians = 90^{o}
theta = Pi/2 radians = 90^{o}



h_{a} = a
p = q = a sqrt(2)
P = 4a
s = 2a
K = a^{2}
R = a sqrt(2)/2
r = a/2 


JavaSketchpad exploration: Square

Rectangle 



A quadrilateral with adjacent sides perpendicular
(all four angles are therefore right angles).
a = c, b = d.
A = B = C = D = Pi/2 radians = 90^{o}



h_{a} = b
h_{b} = a
p = q = sqrt(a^{2}+b^{2})
theta = 2 arctan(a/b)
P = 2(a+b)
s = a + b
K = ab
R = p/2 = sqrt(a^{2}+b^{2})/2
r = minimum(a,b)/2 


JavaSketchpad exploration: Rectangle

Parallelogram 


A quadrilateral with opposite sides parallel.
a = c, b = d
A = C, B = D




A + B = Pi radians = 180^{o}
h_{a} = b sin(A) = b cos(BPi/2)
h_{b} = a sin(A) = a cos(BPi/2)
p = sqrt[a^{2}+b^{2}2ab cos(A)]
q = sqrt[a^{2}+b^{2}2ab cos(B)]
p^{2}+q^{2} = 2(a^{2}+b^{2})
theta = arccos([a^{2}b^{2}]/pq)
P = 2*(a+b)
s = a + b
K = ab sin(A) = ab sin(B) = bh_{b}
= pq sin(theta)/2 


JavaSketchpad exploration: Parallelogram

Rhombus 


A parallelogram with all sides equal.
a = b = c = d
A = C, B = D
theta = Pi/2 radians = 90^{o} 


A + B = Pi radians = 180^{o}
h_{a} = a sin(A) = a cos(BPi/2)
h_{a} = h_{b}
p = a sqrt[22 cos(A)]
q = a sqrt[22 cos(B)]
p^{2}+q^{2} = 4a^{2}
P = 4a
s = 2a
K = a^{2}sin(A) = a^{2}sin(B)
= ah_{a} = pq/2 


JavaSketchpad exploration: Rhombus 
Trapezoid (American)
Trapezium (British)*



a parallel to c, m = (a+c)/2
A + B = C + D = Pi radians = 180^{o}



P = a + b + c + d
K = h_{a}m = h_{a}(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



h_{a}^{2} = (a+bc+d)(a+b+c+d)(abc+d)(a+bcd)/[4(ac)^{2}].


If h_{a}^{2} < 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 IndoEuropean 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

Kite 


A quadrilateral with two pairs of distinct
adjacent sides equal in length.
a = b, c = d
theta = Pi/2 radians = 90^{o}



OB = OD = p/2, OA = h, OC = q  h
h = sqrt(a^{2}p^{2}/4)
q = sqrt(a^{2}p^{2}/4) + sqrt(c^{2}p^{2}/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 = 180^{o}



K = sqrt[(sa)(sb)(sc)(sd)]
(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) /
(sa)(sb)(sc)(sd)]/4
theta = arcsin[2K/(ac+bd)] 
CyclicInscriptable 


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 = 180^{o}
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/(Rm)^{2} = 1/r^{2} 
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