Area of Octagon – formula and solved examples

Area of octagon

What is an octagon?

An octagon is a polygon having eight sides. It can be further classified into two types based on the length of the sides.

1. Regular Octagon

• All sides are equal.
• All interior angles are equal and the measure of each angle is equal to 135°.

2. Irregular Octagon

• All sides are not equal.
• Interior angles are not equal.

The sum of all interior angles of an octagon is equal to 1080°.

Area of a regular octagon

Given below is a regular octagon of side ‘s’ divided into eight equivalent triangles.

So, the area of the octagon is equal to 8 times the area of each triangle.

The above figure shows one of the triangles which is a part of the octagon.

In △ABC

1. AB = AC

2. BC = s

3. ∠BAC = \frac{360^{\circ}}{8}=45^{\circ}

4. AD is a perpendicular bisector of BC.

⇒ ∠ADC = 90° 

\mathrm{BD}=\mathrm{DC}=\frac{s}{2}

5. AD also bisects ∠BAC.

Now, In △DAC 

1. ∠ADC = 90°\angle \mathrm{ADC}=90^{\circ}

2. \angle \mathrm{DAC}=\frac{45^{\circ}}{2}

3. \tan \angle \mathrm{DAC}=\tan \left(\frac{45^{\circ}}{2}\right)=\frac{D C}{A D}

     \Rightarrow \quad A D=\frac{D C}{\tan \frac{45^{\circ}}{2}}

According to trigonometry identities

\tan \frac{\theta}{2}=\frac{\sin \theta}{1+\cos \theta}

\tan \frac{45^{\circ}}{2}=\frac{\sin 45^{\circ}}{1+\cos 45^{\circ}}

=\frac{1}{1+\sqrt{2}}

Substituting the value of tan \frac{45^{\circ}}{2} and DC in the equation

\mathrm{AD}=\frac{D C}{\tan \frac{45}{2}}=\frac{s / 2}{\frac{1}{1+\sqrt{2}}}=\frac{\sqrt{2}+1}{2} \mathrm{~s}

\text { Area of } \triangle \mathrm{ABC}=\frac{1}{2} \times A D \times B C

                                =\frac{1}{2} \times \frac{\sqrt{2}+1}{2} s \times s

=\frac{\sqrt{2}+1}{4} s^{2}

Area of Octagon = 8 × Area of △ABC 

                              = 8 \times \frac{\sqrt{2}+1}{4} s^{2}

                              = 2(\sqrt{2}+1) s^{2}

Hence the area of a regular octagon is 2(\sqrt{2}+1) s^{2}, where s is the length of each side. 

Area of an Irregular octagon

There is no defined formula for finding the area of an irregular octagon.

In this case, we have to divide the octagon into different polygons according to the question and then find the area of the irregular octagon by adding the area of the different polygons.

Solved examples

1. The length of each side of a regular octagon is equal to 15 cm. Find the area of this octagon?

Solution:

Length of each side = s = 15 cm

Using the area formula 

\text { Area }= 2(\sqrt{2}+1) s^{2}

=2(\sqrt{2}+1) 15^{2}

=1086.3961 \mathrm{~cm}^{2}

Hence the area of this octagon is equal to 1086.3961 \mathrm{~cm}^{2}

2. The area of octagon having equal sides is equal to 82 cm². Find the length of each side?

Solution:

\text { Area }=2(\sqrt{2}+1) s^{2}

\Rightarrow 82=2(\sqrt{2}+1) s^{2}

\Rightarrow s^{2}=\frac{82}{2(\sqrt{2}+1)}=16.98

\Rightarrow s=\sqrt{16.98}=4.12 \mathrm{~cm}

Hence the length of each side is equal to 4.12 cm.