Area of an Octagon Formula: Definition, Classification, Examples

An eight-sided two-dimensional geometrical figure is called an octagon. It is derived from the Greek words octa, which meant eight, and gonon, which meant angles. The area of an octagon is defined as the region occupied inside the boundary of an octagon. The area of an octagon formula is represented as \(2 a^{2}(\sqrt{2}+1)\).

There are \(8\) interior angles and \(8\) exterior angles in an octagon. The sum of interior angles in an octagon is \(1080^{\circ}\). In this article, we will learn about the definition of the octagon, properties of an octagon, different types of octagon and formulas to calculate the area and perimeter of a regular octagon.

What is an Octagon?

An octagon is an eight-sided polygon with eight straight lines and eight interior angles, which add up to \(1080^{\circ}\). An octagon shape is a flat (two-dimensional) \(8\)- sided geometric shape. In an octagon, each interior angle measures \(135^{\circ}\), and each exterior angle measures \(45^{\circ}\).

The meaning of octagon shape is derived from the Greek word as octa denotes eight, and gonia denote angle. The octagon is an \(8\) -sided polygon. The best real-life example of an octagon is a stop signboard.

Types of Octagon

Polygons are of four different types: concave polygon, convex polygon, regular polygon, and irregular polygon. Similarly, the octagon has four types. They are,

1. Concave octagon
2. Convex octagon
3. Regular octagon
4. Irregular octagon

1. Concave Octagon

An octagon in which at least one angle is more than \(180^\circ \) is called a concave octagon. In other words, if the vertices point inwards or pointing inside in an octagon, it is known as a concave octagon.

2. Convex Octagon

An octagon in which each angle is less than \(180^{\circ}\) is called a convex octagon. In other words, if the vertices point outwards in an octagon or pointing outside is known as a convex octagon.

3. Regular Octagon

The octagon having eight equal sides and angles is known as a regular octagon. Every regular octagon has the same angle measures. An octagon with all sides equal and each one of the angles equal is called a regular octagon.

4. Irregular Octagon

An octagon that does not have any equal sides or equal angles is an irregular octagon. In an irregular octagon, we have eight unequal sides and eight unequal angles. Octagons that aren’t regular are called irregular octagons. In other words, an octagon in which the sides and angles vary is known as an irregular octagon.

Interior and Exterior Angles of a Regular Octagon

We know that, for a regular polygon of \(n\) sides, we have:

(i) The sum of exterior angles \(=360^{\circ}\)
(ii) Each exterior angle \(=\frac{360^{\circ}}{n}\)
(iii) The sum of interior angles \(=(n-2) \times 180^{\circ}\)
(iv) Each interior angle \(=\frac{(n-2) \times 180^{\circ}}{n}\)

So, for a regular octagon,

The sum of exterior angles \(=360^{\circ}\)

Each exterior angle \(=\frac{360^{\circ}}{8}=45^{\circ}\)

The sum of interior angles \(=(8-2) \times 180^{\circ}=6 \times 180^{\circ}=1080^{\circ}\)

The interior angle of a regular octagon \(=\frac{6 \times 180^{\circ}}{8}=135^{\circ}\)

Perimeter of an Octagon

The computation of the length of the boundary of any closed figure is known as its perimeter. The perimeter of a regular or irregular octagon will be the sum of the lengths of its sides.

If \(a, b, c, d, e, f, g, h\) are respectively the lengths of sides of an irregular octagon, then the perimeter of an irregular octagon is \(a+b+c+d+e+f+g+h\).

Perimeter \(=a+b+c+d+e+f+g+h\)

If \(a\) represents the side of a regular octagon, then the perimeter of a regular octagon \(=8a \)

Number of Diagonals in an Octagon

If the number of sides of the polygon is \(n\) then the number of diagonals we can draw is given by \(\frac{n(n-3)}{2}\).

So, number of diagonals in an octagon \(=\frac{8(8-3)}{2}=20\)

Therefore, the number of diagonals in an octagon is \(20\).

Length of the Diagonal in an Octagon

If \(L\) is the length of the diagonal of an octagon and \(a\) is the length of each side of the octagon, then the length of the diagonal is given by:

\(L=a \sqrt{4+2 \sqrt{2}}\)

Area of an Octagon

The area of an octagon is the region occupied inside the boundary of an octagon.

To calculate the area of a regular octagon, we divide it into small eight isosceles triangles. Compute the area of one of the triangles, and then we can multiply by \(8\) to find the total area of the octagon.

By taking one of the triangles and drawing a line from the apex to the midpoint of the base, it forms a right angle.

The triangle’s base is \(a\), which is the side length of the polygon, and \(OC\) is the triangle’s height.

We know that, \(2 \sin ^{2} \theta=1-\cos 2 \theta…….(i)\)

\(2 \cos ^{2} \theta=1+\cos 2 \theta……..(ii)\)

Dividing \((i)\) and \((ii)\), we get

\(\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}=\frac{1-\cos 2 \theta}{1+\cos 2 \theta}\)

We know that, \(\frac{\sin \theta}{\cos \theta}=\tan \theta\)

So, \(\tan ^{2} \theta=\frac{1-\cos 2 \theta}{1+\cos 2 \theta}\)
\( \Rightarrow {\tan ^2}\left( {\frac{{45}}{2}} \right) = \frac{{1 – \cos \,{{45}^{\rm{o}}}}}{{1 + \cos \,{{45}^{\rm{o}}}}}\)
\(\Rightarrow \tan ^{2}\left(\frac{45}{2}\right)=\frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}\)
\(\Rightarrow \tan ^{2}\left(\frac{45}{2}\right)=\frac{\sqrt{2}-1}{\sqrt{2}+1}\)

Rationalising the denominator, we get
\(\tan ^{2}\left(\frac{45}{2}\right)=\frac{\sqrt{2}-1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}\)
\(\Rightarrow \tan ^{2}\left(\frac{45}{2}\right)=\frac{(\sqrt{2}-1)^{2}}{(\sqrt{2})^{2}-1^{2}}\)
\(\Rightarrow \tan ^{2}\left(\frac{45}{2}\right)=\frac{(\sqrt{2}-1)^{2}}{2-1}\)
\(\Rightarrow \tan \left(\frac{45}{2}\right)=\sqrt{2}-1\)

Since, \(\angle A O C=\angle B O C=\frac{45}{2}\), we can write \(\tan \left(\frac{45}{2}\right)=\frac{A C}{O C}\)

Therefore, \(\frac{A C}{O C}=\sqrt{2}-1\)
\(\Rightarrow O C=\frac{\frac{a}{2}}{\sqrt{2}-1}\)

Rationalising the denominator, we get
\(O C=\frac{a}{2}(\sqrt{2}+1)\)

Now, the area of triangle \(O A B=\frac{1}{2} \times A B \times O C\)
\(=\frac{1}{2} \times a \times \frac{a}{2}(\sqrt{2}+1)\)
\(=\frac{a^{2}}{4}(\sqrt{2}+1)\)

So, the area of the octagon \(=8 \times \frac{a^{2}}{4}(\sqrt{2}+1)\)
\(=2 a^{2}(\sqrt{2}+1)\)

Therefore, we can write the formula for the area of an octagon as \(2 a^{2}(\sqrt{2}+1)\).

Solved Examples on Area of an Octagon Formula

Q.1. Raju was solving a Math sum that required him to calculate the length of each side of a regular octagon while the perimeter of the octagon was given as \(240\) units. Help Raju to find the length of a regular octagon.
Ans: We know that, in a regular octagon, all sides are of equal length. So, let the length of each side of a regular octagon be \(a\).
Also, it is given that the perimeter of a regular octagon is \(240\) units.
We know that if \(a\) represents the side of a regular octagon, then the perimeter of a regular octagon \(=8 a\).
So, \(8 a=240\)
\(\Rightarrow a=\frac{240}{8}=30\)
Therefore, the length of each side of the given regular octagon is \(30) units.

Q.2. Priya is planning to construct an octagonal-shaped garden in front of her house. As per her plan, each side of the octagonal garden will measure \(8\) units. She needs to buy a fence wire to place around the boundary of the park. Find the length of the fence wire required.
Ans: It is given that Priya wants to construct an octagonal-shaped garden that has each side measure \(=8\) units
Also, the length of fence wire required \(=\) perimeter of the octagonal-shaped garden
We know that if \(a\) represents the side of a regular octagon, then the perimeter of a regular octagon \(=8 a\)
So, the length of fence wire required \(=8 \times 8\) units
\(=64\) units.
So, the length of the fence wire required for the Priya \(=64\) units

Q.3. Find the length of the longest diagonal of a regular octagon whose side length is equal to \(10\,\text {cm}\).
Ans: We know that the length of the diagonal of an octagon is \(L=a \sqrt{4+2 \sqrt{2}}\)
Given: \(a=10 \mathrm{~cm}\)
So, length of the diagonal \(=10 \times \sqrt{4+2 \sqrt{2}}\)
\(=10 \times 2.6131 \mathrm{~cm}\)
\(=26.131 \mathrm{~cm}\)
Therefore, the length of the diagonal is \(26.131 \mathrm{~cm}\)

Q.4. Find the perimeter and area of a regular octagon whose side measures \(6 \mathrm{~cm}\).
Ans: We know that if \(a\) represents the side of a regular octagon, then the perimeter of a regular octagon \(=8 a\) and the area of an octagon is \(2 a^{2}(\sqrt{2}+1)\).
Given: \(a=6 \mathrm{~cm}\)
So, the perimeter of the given octagon \(=8 \times 6=48 \mathrm{~cm}\)
Area of the given octagon \(=2 \times 6^{2}(\sqrt{2}+1)\)
\(=72(\sqrt{2}+1)\)
\(=72 \times 2.414\)
\(=173.823 \mathrm{~cm}^{2}\)
Therefore, the perimeter of the given regular octagon is \(48 \mathrm{~cm}\), and the area is \(173.823 \mathrm{~cm}^{2}\)

Q.5. If the area of an octagonal boxing ring is \(4000 \mathrm{~cm}^{2}\). Find the length of each side of that octagonal ring.
Ans: We know that the area of an octagon is \(2 a^{2}(\sqrt{2}+1)\).
Given: area \(=4000 \mathrm{~cm}^{2}\)
So, \(4000=2 a^{2}(\sqrt{2}+1)\)
\(\Rightarrow a^{2}=\frac{4000}{4.828} \mathrm{~cm}\)
\(\Rightarrow a^{2}=828.5 \mathrm{~cm}\)
\(\Rightarrow a=28.78 \mathrm{~cm}\)
Therefore, each side of an octagonal boxing ring measures \(28.78 \mathrm{~cm}\).

Summary

An octagon is defined as an eight-sided polygon with eight straight lines and eight interior angles. The octagon can be classified into four types. The four types of octagons are concave, convex, regular, and irregular octagons. The area of an octagon is calculated by the formula \(2 a^{2}(\sqrt{2}+1)\). Furthermore, it is important to note that the number of diagonals in an octagon is 20. The length of the diagonal for the octagon is calculated by the formula \(L=a \sqrt{4+2 \sqrt{2}}\).

FAQs on Area of an Octagon Formula

Q.1. Is an octagon a quadrilateral?
Ans: No, a quadrilateral is a closed polygon with four vertices and four sides. An octagon is composed of eight sides. So, an octagon cannot be a quadrilateral.

Q.2. How many diagonals can we draw in an octagon?
Ans: If the number of sides of the polygon is \(n\) then the number of diagonals we can draw is given by \(\frac{n(n-3)}{2}\)
So, number of diagonals in an octagon \(=\frac{8(8-3)}{2}=20\).

Q.3. How do you find the area of an octagon?
Ans: We can write the formula for the area of an octagon as \(2 a^{2}(\sqrt{2}+1)\), where \(a\) is the length of each side of a regular octagon.

Q.4. What are concave and convex octagons?
Ans: An octagon in which each angle is less than \(180^{\circ}\) is called a convex octagon.
An octagon in which at least one angle is more than \(180^{\circ}\) is called a concave polygon.

Q.5. What is the formula to find the perimeter of an irregular octagon?
Ans: If \(a, b, c, d, e, f, g, h\) are respectively the lengths of sides of an irregular octagon, then the perimeter of an irregular octagon is \(a+b+c+d+e+f+g+h\).

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