Question

How many possible formulas are there for quadrilateral?

Answer 1

Answer:

9

Step-by-step explanation:

The opposite sides of a parallelogram are equal in length. Thus AB = DC and AD = BC.

Opposite angles of a parallelogram are congruent (equal measure).

The diagonals of a parallelogram bisect each other. Here BE = ED and AE = EC.

Each diagonal of a parallelogram divides it into two triangles of the same area. Here area of  ΔABC = Area of ΔACD and area of ΔABD = area of ΔBCD.

Bisectors of the angles of parallelogram form a rectangle.

A parallelogram inscribed in circle is a rectangle.

A parallelogram circumradius about a circle is a rhombus.

The sum of the squares of the diagonals is equal to the sum of the square of the four sides.

Here AC2 + BD2 = AB2 + BC2 + CD2 + DA2 = 2 b2 + 2 c2

Answer 2

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: 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

General




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
cos2([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 = 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



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 = 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

Parallelogram

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

Rhombus



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

Kite



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

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) /
(s-a)(s-b)(s-c)(s-d)]/4

theta = arcsin[2K/(ac+bd)]

Cyclic-Inscriptable



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..