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Area of Polygons – Methods, Formulas & Examples

area of polygon

The area of a polygon is the region enclosed by the boundary(perimeter) of the polygon. The unit of the area of any polygon is always expressed in square units such as $cm^{2}$, $m^{2}$, $in^{2}$, $ft^{2}$, etc. There are pre-defined formulas for computing the area of rectangle, square, triangle, circle, rhombus, trapezium, and kite. However, for computing an area of a polygon, it is calculated by viewing it as a combination of two or more regular polygons

Let’s learn how to find the area of polygons.

Polygon – A 2D Plane Figure

A polygon is a  two-dimensional closed shape bounded with straight sides known as its sides or edges. It does not have curved sides. The points where two sides meet are the vertices (singular vertex) of a polygon. Each vertex contains two sides forming an angle.

area of polygon

The polygons are named based on the number of sides present in them. Some of the common polygons are

Number of SidesName of the Polygon
$3$Triangle (Prefix is Tri meaning $3$)
$4$Quadrilateral (Prefix is Quadri meaning $4$)
$5$Pentagon (Prefix is Penta meaning $5$)
$6$Hexagon (Prefix is Hexa meaning $6$)
$7$Heptagon (Prefix is Hepta meaning $7$)
$8$Octagon (Prefix is Octa meaning $8$)
$9$Nonagon (Prefix is Nona meaning $9$)
$10$Decagon (Prefix is Deca meaning $10$)

Note: 

  • The suffix ‘gon’ in a polygon is a Greek word meaning side
  • The ‘poly’ is also a Greek word meaning many

Regular and Irregular Polygons

A polygon is broadly categorized as a regular and irregular polygon based on the length of its sides. As the name suggests the length of all sides is equal and also the measure of each angle is equal. Therefore, one can use a fixed or a general formula to find the perimeter of a polygon by using either of the following

  • number of sides
  • measure of an interior angle
  • measure of an exterior angle

In the case of an irregular polygon, the length of the sides is different, and also the measure of angles is different. Therefore, to find the perimeter of an irregular polygon length of all the sides should be known.

Area of Polygon – Regular

Before moving on to the formula for finding the area of a polygon, let’s understand some terms associated with it.

Centre of a Regular Polygon: The centre of a regular polygon is the point from which all the vertices are equidistant. 

Radius of a Regular Polygon: The radius of a regular polygon is a segment with one endpoint at the centre and the other endpoint at one of the vertices. Thus, there are $n$ radii in an $n$-sided regular polygon.

area of polygon

Note: The centre and radius of a regular polygon are the same as the centre and radius of a circle circumscribed about that regular polygon.

Apothem of a Regular Polygon: An apothem of a regular polygon is a segment with one endpoint at the centre and the other endpoint at the midpoint of one of the sides. The apothem of a regular polygon is the perpendicular bisector of whichever side on which it has its endpoint.

area of polygon

Central Angle of a Regular Polygon: A central angle of a regular polygon is an angle whose vertex is the center and whose rays, or sides, contain the endpoints of a side of the regular polygon.

area of polygon

Note: An $n$-sided regular polygon has $n$ apothems and $n$ central angles, each of whose measure is $\frac {360}{n}$ degrees. Every apothem is the angle bisector of the central angle that contains the side to which the apothem extends.

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Maths can be really interesting for kids

Finding Length of Apothem

Consider a pentagon with centre $O$ and length of each side $s$. Further, consider a $\triangle OAB$ on one of the sides $AB$. (There will be $5$ such triangles in a pentagon).

area of polygon

The measure of central angle $\theta = \frac {360}{n}$ degrees (In case of pentagon, $n = 5$).

Note: 

  • $OC$ is the apothem of the pentagon
  • $\triangle OAB$ is an isosceles triangle, where $OA = OB$ (radii of a polygon)
  • The apothem $OC$ bisects $\angle AOB$, i.e., $\angle AOC = \angle COB = \frac {\theta}{2} = \frac {180}{n} \text {degrees}$

In $\triangle AOC$, $\tan \frac {\theta}{2} = \frac {\frac {s}{2}}{a} => a = \frac {\frac {s}{2}}{\tan \frac{\theta}{2}}$

The area of a regular polygon is one-half the product of its apothem and its perimeter, i.e., $\text {Area} = \frac {1}{2} \times a \times p$, where $a$ is the length of an apothem, and $p$ is the perimeter.

Therefore, formula for area of a regular polygon is $\text {Area} = \frac {1}{2} \times \frac {\frac {s}{2}}{\tan \frac{\theta}{2}} \times p$.

The perimeter of a $n$-sided polygon is given by $p = n \times s$

Therefore, $\text {Area} = \frac {1}{2} \times \frac {\frac {s}{2}}{\tan \frac{\theta}{2}} \times n \times s = \frac {ns^{2}}{4 \tan \frac {\theta}{2}}$.

Substituting the value of $\frac {\theta}{2} = \frac {180}{n}$ in the formula, finally we get

$\text{Area} = \frac {ns^{2}}{4 \tan \frac {180}{n}}$

Let’s now see how $\frac {1}{2} \times a \times p$ is equal to the sum of the areas of the triangles that make up a regular polygon. 

Let’s consider a pentagon (a $5$-sided figure). It can be divided into $5$ triangles. One such triangle is shown in the figure below.

area of polygon

Area of triangle is $\frac {1}{2} \times s \times a$.

Area of pentagon shown in the figure is sum of area of $5$ triangles, i.e., $\text {Area of Pentagon} = 5 \times \frac {1}{2} \times s \times a = \frac {1}{2} \times \left(5 \times s \right) \times a = \frac {1}{2} \times p \times a$ (where $p$ is the perimeter of pentagon).

Note: When an $n$-sided polygon is split up into $n$ triangles, its area is equal to the sum of the areas of the triangles.

Area Formula for Different Polygons

Regular PolygonNumber of sides $n$Area Formula
Equilateral Triangle$3$$\frac {3s^{2}}{4 \tan 60^{\circ}}$
Square$4$$\frac {ns^{2}}{4 \tan 45^{\circ}}$
Regular Pentagon$5$$\frac {ns^{2}}{4 \tan 36^{\circ}}$
Regular Hexagon$6$$\frac {ns^{2}}{4 \tan 30^{\circ}}$
Regular Heptagon$7$$\frac {ns^{2}}{4 \tan 25.71^{\circ}}$
Regular Octagon$8$$\frac {ns^{2}}{4 \tan 22.5^{\circ}}$
Regular Nonagon$9$$\frac {ns^{2}}{4 \tan 20^{\circ}}$
Regular Decagon$10$$\frac {ns^{2}}{4 \tan 18^{\circ}}$

Examples

Ex 1: Find the area of a regular octagon whose length of each side is $12 cm$.

For a regular octagon number of sides $n = 8$

Therefore, $\theta = \frac {360}{8} = 45^{\circ}$ and $\frac {\theta}{2} = 22.5^{\circ}$

Length of each side $s = 12 cm$

$\text{Area} = \frac {ns^{2}}{4 \tan \frac {180}{n}} = \frac {8 \times 12^{2}}{4 \tan 22.5^{\circ}} = \frac {1152}{4 \times (\sqrt {2} – 1)}$ 

$= \frac {288}{\sqrt {2} – 1} = \frac {288}{\sqrt {2} – 1} \times \frac {\sqrt{2} + 1}{\sqrt{2} + 1} = \frac {288 \left(\sqrt{2} + 1 \right)}{\left(\sqrt{2} \right)^{2} – 1^{2}}$

$ = \frac {288 \left(\sqrt{2} + 1 \right)}{2 – 1} = 288\left(\sqrt{2} + 1 \right) cm^{2}$

Ex 2: Find the area of a square whose length of each side is $10 in$.

Note: A square is a regular quadrilateral

Number of sides $n = 4$

Therefore, $\theta = \frac {360}{4} = 90^{\circ}$ and $\frac {\theta}{2} = 45^{\circ}$

Length of each side $s = 10 cm$

$\text{Area} = \frac {ns^{2}}{4 \tan \frac {180}{n}} =  \frac {4 \times 10^{2}}{4 \tan 45^{\circ}} = \frac {4 \times 10^{2}}{4 \times 1} = 100 in^{2}$

Proper & Improper Fractions

Area of Polygon – Irregular

An irregular polygon is a plane-closed shape that does not have equal sides and equal angles. Thus, in order to calculate the area of irregular polygons, we split the irregular polygon into a set of regular polygons such that the formulas for their areas are known. 

Consider the polygon (irregular quadrilateral) shown below.

area of polygon

Divide the quadrilateral into two triangles by joining the vertices $B$ and $D$. (Drawing a diagonal $BD$).

Now, you can use the area of triangle formula to calculate the areas of $\triangle ABD$ and $\triangle BCD$ and calculate the area of irregular quadrilateral $ABCD$ by adding the areas of two triangles.

Area of quadrilateral $ABCD = $ (Area of $\triangle ABD$) + (Area of $\triangle BCD$).

Area of Polygons With Coordinates

To find the area of polygons with known coordinates, the first step is to find the length of each side of the polygon using the distance formula.

The distance (or length) between two points $\left(x_{1}, y_{1} \right)$ and $\left(x_{2}, y_{2} \right)$ is given by the formula $D = \sqrt{\left(x_{2} – x_{1} \right)^{2} + \left(y_{2} – y_{1}\right)^{2}}$.

Now do one of the following

  • If the polygon is a regular polygon then find the area using the formula.
  • If the polygon is an irregular polygon then find the area by partitioning it into a number of smaller polygons such as triangles, squares, rectangles, etc.

Examples

Ex 1: Find the area of a quadrilateral with vertices $\left(4, 10 \right)$, $\left(9, 7 \right)$, $\left(11, 2 \right)$, and $\left(2, 2 \right)$.

The four vertices of a quadrilateral are $A\left(4, 10 \right)$, $B\left(9, 7 \right)$, $C\left(11, 2 \right)$, and $D\left(2, 2 \right)$.

area of polygon

Length of side $AB = \sqrt{\left(9 – 4 \right)^{2} + \left(7 – 10 \right)^{2}} = \sqrt{5^{2} + \left(-3 \right)^{2}} = \sqrt{25 + 9} =\sqrt{34} = 5.83$ units

Length of side $BC = \sqrt{(11 – 9)^{2} + (2 – 7)^{2}} = \sqrt{2^{2} + \left(-5 \right)^{2}} = \sqrt{4 + 25} = \sqrt{29} = 5.39$ units

Length of side $CD = \sqrt{\left(2 – 11 \right)^{2} + \left(2 – 2 \right)^{2}} = \sqrt{\left(-9 \right)^{2} + 0^{2}} = \sqrt{81} =9$ units

Length of side $DA = \sqrt{\left(4 – 2 \right)^{2} + \left(10 – 2 \right)^{2}} = \sqrt{2^{2} + 8^{2}} = \sqrt{4 + 64} = \sqrt{68} = 8.25$ units

Length of diagonal $AC = \sqrt{\left(11 – 4 \right)^{2} + \left(2 – 10 \right)^{2}} = \sqrt{7^{2} + \left(-8 \right)^{2}} = \sqrt{49 + 64} = \sqrt{113} = 10.63$ units

To find the area of triangles we’ll use Heron’s formula.

Area of $\triangle ABC$

$a = AB = 5.83$ units, $b = BC = 5.39$ units and $c = AC = 10.63$ units

Semiperimeter $s = \frac {a + b + c}{2} = \frac {5.83 + 5.39 + 10.63}{2} = 10.92$ units

Area = $\sqrt{s \left(s – a \right)\left(s – b \right)\left(s – c \right)} = \sqrt{10.92 \left(10.92 – 5.83 \right)\left(10.92 – 5.39 \right)\left(10.92 – 10.63 \right)} = 9.44 units^{2}$

Area of $\triangle ACD$

$a = AC = 10.63$ units, $b = CD = 9$ units and $c = DA = 8.25$ units

Semiperimeter $s = \frac {a + b + c}{2} = \frac {10.63 + 9 + 8.25}{2} = 13.94$ units

Area = $\sqrt{s \left(s – a \right)\left(s – b \right)\left(s – c \right)} = \sqrt{13.94 \left(13.94 – 10.63 \right)\left(13.94 – 9 \right)\left(13.94 – 8.25 \right)} = 36.01 units^{2}$

Therefore, area of quadrilateral $ABCD = 9.44 + 36.01 = 45.45 units^{2}$.

Conclusion

A polygon is a  two-dimensional closed shape bounded with straight sides known as its sides or edges. A polygon can be a regular or irregular polygon. You can find the area of a regular polygon by using the pre-defined formula, whereas to find the area of an irregular polygon, first of all, divide the polygon into a number of polygons whose area formula are known and then find the area of an irregular polygon by adding the areas of all polygons so formed.

Practice Problems

  1. Find the area of an equilateral triangle of side $7 cm$.
  2. Find the area of a square of side $9 mm$.
  3. Find the area of a regular pentagon of side $11 cm$.
  4. Find the area of a regular hexagon of side $5 m$.
  5. Find the area of a regular heptagon of side $7 cm$.
  6. Find the area of a regular octagon of side $4 in$.
  7. Find the area of a regular nonagon of side $12 cm$.
  8. Find the area of a regular decagon of side $6 cm$.

Recommended Reading

FAQs

What is the definition of the area of a polygon?

The area of a polygon is the region occupied by a polygon. Polygons can be regular and irregular. The basic polygons which are used in geometry are triangle, square, rectangle, pentagon, hexagon, etc. 

What is the formula for the area of a regular polygon?

The formula for the area of a $n$-sided polygon is given by $\text{Area} = \frac {ns^{2}}{4 \tan \frac {180}{n}}$, where $s$ is the side length.

How do you find the area of an irregular polygon?

To calculate the area of irregular polygons, we split the irregular polygon into a set of regular polygons such that the formulas for their areas are known.

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