Area of Polygon

A Polygon is generally explained as a closed figure with its all sides (or edges) with straight lines.
Polygon gets its name from the combination of two Greek words – poly-gon and the meaning is many angles or corners.
The characteristics of the polygon are:
  • At least three sides are needed
  • The figure must be closed
  • The sides must be straight lines
  • Lines intersect in pairs.

For example:

Polygons

The polygons are grouped or named by their number of sides. The smallest polygon is the one with 3 sides called the Triangles,
4 sides are Quadrilaterals,
5 sides are Pentagons,
6 sides are Hexagons,
7 sides are Heptagons,
8 sides are Octagons and goes on.

They are also grouped as regular and irregular polygons. Regular polygons have all sides and angles equal or else it is an irregular polygon.

Area of a Polygon Formula

Triangle: As Polygon is a closed figure, surely it has some area bounded by its boundary.

The general method for finding the area of a Polygon which is applicable to all kind of polygons is to divide the polygon into different triangles, then find the area of each triangle and then sum up the area of these triangles to get the total area of the Polygon.

The general formula for the area of a triangle A = ½bh
Where b is the base length and
h is the height of the triangle.

Let us check this with a square. Square is a special case of quadrilateral with all sides are equal and all angles measure 90° each.
We know that the area of a square is a² where ‘a’ length of its side.
Now if we divide the square through the diagonal as shown, we get two triangles where the area of each triangle is ½ a*a = ½a².

Area of Triangle Polygon
Adding the area of two triangles, that is ½a² + ½a² = a² which is the area of the square.

This method can be used to find the area for the entire polygon.

Area of a rectangle: Rectangle is a quadrilateral with opposite sides equal and each angle equal to 90 degree.
Let l be the length and w be the width of a rectangle,

Area of Rectangle
then area of the rectangle is the product of the its length and the width.
So Area, A =lw
For example, if 4 unit is the length and 2 unit is the width of a rectangle, then its area = 4*2=8 unit²
Parallelogram: Parallelogram is a quadrilateral, with opposite sides equal and parallel.

Area of Parallelogram
 
To find the area draw a perpendicular segment from one vertex of the opposite side of a parallelogram. This can be denoted as h, which is the height of the parallelogram.
 
Solving Area of Parallelogram

So the area of a parallelogram is the product of its height and length
That is A = hl
For example, if the 5 unit is the length and 2.82 unit is the height of the parallelogram, then the area of the parallelogram = 5*2.82 = 14.1 unit²
Rhombus: Rhombus is a quadrilateral with all sides equal and opposite sides is parallel.
 
Rhombus

To find the area of the rhombus, first draw the diagonals of the rhombus, say measure a and b units..
 
Solving Rhombus

Then the area of the rhombus is the half the product of their diagonals
So A = $\frac{1}{2}$ ab
For example, if the diagonals of a rhombus are 2.97 units and 3.28units, then the area of the rhombus = $\frac{1}{2}$*2.97*3.28
                        = 4.8708units²

This same formula is used to find the area of the kite.
Trapezoid: Trapezoid is a quadrilateral, with only one opposite sides are parallel, and the other opposite sides are non parallel.

Trapezoid
 
To find the area, first draw the height of the trapezoid. Which is drawn perpendicular to its parallel side.
 
Solving Trapezoid

Then area of a trapezoid, is the sum of the length of the parallel sides multiplied with half of the height.
Area , A =  $\frac{1}{2}$*(a+b)h

Area of a Regular Polygon

Now, consider a regular polygon, say pentagon as shown below with its length of a side as 2l, and apothem (radius of inscribed circle) as r. Now from the figure we can see that the pentagon can be divided into ten triangles with same area.
 
Area of a Regular Polygon

 So the area of the pentagon is 10 times the area of a triangle.

Let us consider the triangle ABF whose area is ½lr
So the area of the pentagon is 10 x ½lr = 5lr
Here 5 (since we have taken a pentagon) represents the number of sides, l represents ½ the length of a side and r is the apothem (radius of inscribed circle)      
So we can say that the general formula for the area of a regular polygon,
                A = nlr,

where, n represents the number of sides.
Further ‘nl’ is half the perimeter of the regular polygon.  So the area of a regular polygon can be generalized as half perimeter of the regular polygon multiplied by the apothem (radius of inscribed circle).
A = ½pr, where p is the perimeter of the regular polygon and r is the apothem (radius of inscribed circle)  

Application of trigonometry: by applying trigonometry, we can find the ratio of side with the apothem and hence the area of the polygons

Area of a regular or irregular Polygon using the coordinates

Consider a polygon made of n vertices in the form (xi, yi)  , i=0,2,..n-1
There will always be an assumption that the nth vertex which is (xn, yn) will be same as (x0, y0) as the figure is closed.

Hence the general form of finding the area of a polygon where the coordinates of the vertices are given, is
Area =  $\frac{|(x1 \times y2 - y1 \times x2) + ((x2 \times y3 - y2 \times x3) + ………… + (xn \times y1 - yn \times x1)|}{2}$
                                                           
Where, (x1,y1), ((x2,y2), (x3,y3), ………., (xn, yn) are the coordinates of the vertices of the polygon.

Consider a polygon whose coordinates of the vertices are (x1,y1), ((x2,y2), (x3,y3), (x4, y4) as shown

Area of a Irregular PolygonArea of an Irregular Polygon
 
Its area will be
Area =  $\frac{|(x1 \times y2 - y1 \times x2) + ((x2 \times y3 - y2 \times x3) + (x4 \times y1-y4 \times x1)|}{2}$
                                       
                   
For example:
Find the area of the following figure
 
Calculating Polygon Area

Solution:
There are 9 vertices
So area = $\frac{1}{2}$$\sum_{i-0}^{8}$ (xi yi + 1– xi + 1yi)

=$\frac{1}{2}$ [(12*7-7*13)+(13*6-7*13)+(13*8-14*6)+(14*7-8*16)+(16*8-15*7)+(15*10-14*8)+(14*8-13*10)+(13*9-12*8)+(12*7-9*12)]
= $\frac{1}{2}$ [-7+(-13)+20+(-30)+23+38+(-18)+21+(-24)]
= $\frac{1}{2}$ *10
= 5 unit²