General Formula for the Area of Quadrilaterals

Some formulas for area in terms of sides a, b, c, and d, and diagonal lengths e1 and e2 are as follows:

𝑨 =𝟏

𝟐𝒆𝟏𝒆𝟐 𝐬𝐢𝐧 𝜽

where θ is the angle formed between e1 and e2.

𝑨 =𝟏

𝟒𝒂𝟐 + 𝒄𝟐 − 𝒃𝟐 − 𝒅𝟐 𝒕𝒂𝒏𝜽

where the four sides are labeled such that a2+c2 > b2+d2

ab

cd

C

D

A

Be1

e2

θ

General Formula for the Area of Quadrilaterals

𝑨 = 𝒔 − 𝒂 𝒔 − 𝒃 𝒔 − 𝒄 𝒔 − 𝒅 − 𝒂𝒃𝒄𝒅𝒄𝒐𝒔𝟐𝟏

𝟐𝑨 + 𝑪

Where s is the semi perimeter and angles A and C are any two opposite angles of the quadrilateral.

Parallelogram

A parallelogram is a quadrilateral whoseopposite sides are parallel.

A

C

B

D

h (height)

b (base)

Parallelogram

Parallelograms have the followingimportant properties:

1. Opposite sides are equal.2. Opposite interior angles are congruent

( e.g. ∠𝑨 ≅ ∠𝑫).3. Adjacent angles are supplementary (

e.g. ∠𝑨 + ∠𝑪 = 𝟏𝟖𝟎°)4. A diagonal divides the parallelogram

into two congruent triangles ( e.g.Δ𝑪𝑨𝑩 = Δ 𝑪𝑫𝑩)

5. The two diagonals bisect each other.

A

C

B

D

Diagonals of a Parallelogram

A

C

B

D

a

b

d

ha

h

θ

By cosine law:

d2 = a2 + b2 – 2 ab cos θ

If any two parts are given, the relationship among a, h and θ may be obtained from the right triangle as shown.

Using the other angle, 180° - θ the second diagonal may be obtained by the same formula.

Parallelogram

Perimeter of a Parallelogram: P = 2a + 2b

Area of a Parallelogram:

A = bhA = absin θ

where b is the length of the base, h is the height , and b are the sides and θ is any interior angle.

Diagonals of a Rectangle

A

C

B

D

h

b

d = 𝑏2 + ℎ2

Perimeter of a Rectangle

P = 2b + 2h

A

C

B

D

h

b

Area of a Rectangle

A

C

B

D

h

b

A = bh

Diagonals of a Square

d = 𝑎2 + 𝑎2 = 𝑎 2

a

a

d

Perimeter of a Square

P = 4a

a

a

d

Area of a Square

A = a2

a

a

d

Diagonal of a Rhombus

h

Diagonals of rhombus are perpendicular bisectors.Angle between them is 90°.

Using Phytagorean theorem, diagonals may beobtained like in a similar manner like that of aparallelogram.

𝑏 =𝑑12

2

+𝑑22

2

b

Diagonal of a Rhombus

h

Where d1 and d2 are the shorter and longerdiagonals respectively, and θ is the angle opposited1.

𝜃 = 2 𝑡𝑎𝑛−1𝑑1𝑑2

b

Perimeter of a Rhombus

h

P = 4b

b

Area of a Rhombus

h

𝐴 =1

2𝑑1𝑑2

b

𝐴 = bh