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# Area of Polygons â€“ Explanation & Examples

Whenever we talk about geometry, we talk about side lengths, angles, and areas of the shapes. We saw the other two before; letâ€™s talk about the latter. You got to see so many mathematics exam questions regarding finding the shaded region of a particular polygon.

For that, you need to have the knowledge of formulas of area for different kinds of polygons.

**In this article, you will learn:**

- What the area of a polygon isÂ
- How to find the area of a polygon, including the area of a regular and irregular polygon?

## Â

## What is the Area of a Polygon?

In geometry, the area is defined as the region occupied inside the boundary of a two-dimensional figure. Therefore,** the area of a polygon is the total space or region bound by the sides of a polygon.**

The standard units for the measurement of the area are square meters (m^{2}).

## How to Find the Area of a Polygon?

**Regular polygons** such as rectangles, squares, trapeziums, parallelograms, etc., have pre-defined formulas for calculating their areas.

However, for an** irregular polygon**, the area is calculated by subdividing an irregular polygon into small sections of regular polygons.

### Area of a regular polygon

Calculating a regular polygon area can be as simple as finding the area of a regular triangle. Regular polygons have equal side lengths and an equal measure of angles.

There are **three methods of calculating the area of a regular polygon**. Each method is used on different occasions.

#### Area of a polygon using the concept of the apothem

The area of a regular polygon can be calculated using the concept of apothem. The apothem is a line segment that joins the polygon’s center to the midpoint of any side that is perpendicular to that side. Therefore, the area of a regular polygon is given by;

A = 1/2 . p . a

where p = the perimeter of the polygon = sum of all the side lengths of a polygon.

a = apothem.

Consider a pentagon shown below;

If the apothem, a = x and the length of each side of the pentagon is s, then the area of the pentagon is given by;

**Area = 1/2 . p . a**

Perimeter = s + s + s + s + s

= 5s

So, substitution,

Area = (Â½)5sx

= (5/2) (s. x) Sq. units

When using the apothem method, the length of the apothem will always be provided.

#### Area of a polygon using the formula: A =Â (L^{2 }n)/[4 tan (180/n)]

Alternatively, the area of area polygon can be calculated using the following formula;

A =Â (L^{2 }n)/[4 tan (180/n)]

Where, A = area of the polygon,

L = Length of the side

n = Number of sides of the given polygon.

#### Area of a circumscribed polygon

The area of a polygon circumscribed in a circle is given by,

**A = [n/2 Ã— L Ã— âˆš (RÂ² – LÂ²/4)]** square units.

Where n = number of sides.

L =Side length of a polygon

R = Radius of the circumscribed circle.

Letâ€™s work out a few example problems about the area of a regular polygon.

*Example 1*

Find the area of a regular hexagon, each of whose sides measures 6 m.

__Solution__

For a hexagon, the number of sides, n = 6

L = 6 m

A =Â (L^{2}n)/[4tan(180/n)]

By substitution,

A =Â (6^{2 }6)/ [4tan (180/6)]

= (36 * 6)/ [4tan (180/6)]

= 216/ [4tan (180/6)]

= 216/ 2.3094

A = 93.53 m^{2}

*Example 2*

Find the area of a regular hexagon whose apothem is 10âˆš3 cm and the side length are 20 cm each.

__Solution__

**Area = Â½ pa**

First, find the perimeter of the hexagon.

p = (20 + 20 + 20 + 20 + 20 + 20) cm = (20 cm * 6)

= 120 cm

Substitute.

**Area = Â½ pa**

= Â½ *120 * 10âˆš3

= 600âˆš3 cm^{2}

*Example 3*

Find the regular pentagon area if the polygon’s length is 8 m and the radius of the circumscribed circle is 7 m.__Solution__**A = [n/2 Ã— L Ã— âˆš (RÂ² – LÂ²/4)]** square units.

Where, n = 5; L = 8 m and R = 7 m.

By substitution,

A = [5/2 Ã— 8 Ã— âˆš (7Â² – 8Â²/4)] m^{2}

= [20âˆš (49 â€“ 64/4)]

= 20âˆš (49 â€“ 16)

= 20âˆš33 m^{2}

= 20 * 5.745 m^{2}

= 114.89 m^{2}

*Example 4*

Find the area of a regular pentagon whose apothem and side length are 15cm and18 cm, respectively.

__Solution__

Area = Â½ pa

a = 15cm

p = (18 * 5) = 90 cm

A = (Â½ * 90 * 15) cm

= 675 cm.

**Area of an irregular polygon**

An irregular polygon is a polygon with interior angles of different measures. The side lengths of an irregular polygon are also of different measure.

As said before, we can calculate the area of an irregular polygon by subdividing an irregular polygon into small sections of regular polygons.

*Example 5*

Find the area of an irregular polygon shown below if, *AB = ED *= 20 cm, *BC = CD* = 5cm and *AB = BD* = 8 cm

__Solution__

Subdivide the irregular polygon into sections of regular polygons

Therefore, *ABED* is a rectangle, and *BDC* is a triangle.

Area of rectangle = l * w

= 20 * 8 = 160 cm^{2}

Area of the triangle = 1/2 . b . h

The height the triangle can be calculated by applying the Pythagoras theorem. For instance,

c^{2 }= a^{2} + b^{2}

25^{2 }= a^{2} + 4^{2}

a = âˆš (25 â€“ 16)

a = 3

A = Â½bh = Â½ * 3 * 8

= 6 cm^{2}

Now add the partial areas.

Area of polygon = (160 + 6) cm^{2} =166 cm^{2}