area of a polygonal region

area of a polygonal region

We consider only

Euclidean geometry

.

A

triangular region

is a

triangle

together with its interior.
The sides of the triangle are called the

edges

of the region and the vertices of the
triangle are called the

vertices

of the region.

A

polygonal region

is a plane figure that can be written as a union of a finite
number of triangular regions, subject to a constraint. The constraint is that
the triangular regions are nonoverlapping. That means that if two triangular regions

intersect

, then their

intersection

is either a vertex or edge of each of the triangular regions.

A given polygonal region can of course be written as such a union in infinitely many ways.

We wish to assign a number to each polygonal region that corresponds to our intuitive idea of
”area”. To do that we first write down a set of

postulates

that ”area” should

satisfy

and then
show that the postulates can be satisfied. We let

ℛ

denote the set of polygonal regions.

1.

There is a

function

α

:

ℛ

→

ℝ

.

2.

α

>

0

.

3.

If two triangular regions are

congruent

, then they have the same area.

4.

If two polygonal regions

R

1

and

R

2

are nonoverlapping then

α

⁢

(

R

1

∪

R

2

)

=

α

⁢

(

R

1

)

+

α

⁢

(

R

2

)

.

5.

If a square has edges of length 1, then its area is 1.

The fifth postulate is needed because if

α

satisfies the first 4 postulates, then so does

2

⁢

α

. So, this postulate serves to make

α

a unique function.

Theorem 1

.

If a square has sides of length

s

then its area is

s

2

.

Proof.

We first show that if a square has sides of length

1

q

then its area is

1

q

2

. A

unit square

can be decomposed into

q

2

squares, each having side of length

1

q

. Each of the smaller squares has the same area,

A

, by postulates 3 and 4.
Hence,

1

=

q

2

⁢

A

,

so that

A

=

1

q

2

. Next, if a square has sides of length

p

q

then its area
is

p

2

q

2

. This is because the square can be written as

p

2

squares each having a side of length

1

q

. If

A

is the area then

A

=

p

2

⁢

1

q

2

=

p

2

q

2

.

Finally, let

T

be a square of side

s

and

T

p

/

q

denote a square of side

p

q

having
one angle in common with

T

. In the following

sequence

of statements, each statement
is

equivalent

to the next one:

1.

p

q

<

s

.

2.

T

p

/

q

⊂

T

.

3.

α

⁢

(

T

p

/

q

)

<

α

⁢

(

T

)

.

4.

p

2

q

2

<

α

⁢

(

T

)

.

5.

p

q

<

α

⁢

(

T

)

.

Hence, statements 1 and 5 are equivalent so that

s

=

α

⁢

(

T

)

,

so that

s

2

=

α

⁢

(

T

)

.
This proves the

theorem

.

One can then proceed on the basis of the postulates to show the following sequence of results.

Theorem 2

.

The area of a rectangular region is the

product

of its base and its

altitude

.

Rather than speak of the area of a triangular region or the area of a

rectangle

region,
we shall refer to the area of a triangle, or area of a rectangle, or other polygonal
region, as is usually done.

Theorem 3

.

The area of a

right triangle

is 1/2 product of the lengths of its legs.

Theorem 4

.

The area of a triangle is 1/2 the product of any base and the corresponding altitude.

Theorem 5

.

The area of a

parallelogram

is the product of a base and the corresponding altitude.

Theorem 6

.

The area of a

trapezoid

is 1/2 the product of the altitude and the sum of the bases.

Theorem 7

.

If two triangles are

similar

then the ratio of their areas is the square of the ratio
of any two corresponding sides.

What remains to do is to show that there is a function

α

that satisfies postulates
1 to 5.
It might seem reasonable to just define the area of a rectangle, but rectangles are not
useful for dissecting polygonal regions. The easier way to use triangles.
So one

defines

the area of a triangle to be
1/2 the product of the base and its corresponding altitude. For this to be

well-defined

one has to show first that the product of a base and altitude (for a given triangle)
does not depend on which base is chosen. Next, one defines the

area of a polygonal region

as the sum of the areas of the triangular regions in some decomposition. Of course one has to
show that this sum does not depend on the particular decomposition that is used.
Given this

definition

for

α

, it is then a simple matter to show that
postulates 1 to 5 are satisfied.

The function

α

can be defined on a larger domain, which includes

circular sectors

and disks.
But to do this rigorously ones needs the

concept

of a limit.
In the

hyperbolic geometry

case, one can show the following theorem.

Theorem 8

.

If

α

is a function that satisfies postulates 1 to 4 then there is a

k

>

0

such
that

α

⁢

(

R

)

=

k

⁢

δ

⁢

(

R

)

for every polygonal region, where

δ

⁢

(

R

)

is the defect of

R

.

References

1

Edwin E. Moise,

Elementary

Geometry

from an Advanced Standpoint, Third edition

,
Addison-Wesley, 1990

Title

area of a polygonal region

Canonical

name

AreaOfAPolygonalRegion

Date of creation

2013-03-22 17:00:29

Last modified on

2013-03-22 17:00:29

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

12

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 51M25

Related topic

Area2

Related topic

InterceptTheorem

Related topic

BaseAndHeightOfTriangle

Defines

triangular region

Defines

polygonal region

Defines

area of a triangle

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