Area of regular octagon can be found the following way.

From the center of the octagon we can draw lines to each of the vertex dividing the octagon into 8 triangles.

We find the area of one triangle and multiply the result by 8. The central angle is divided into 8 equal parts. Each angle would be

$\frac{2 \pi}{8}$ or

$\frac{\pi}{4}$ degrees. We drop a perpendicular from the center to the base. This line is called the apothem.

Observe the triangle shown the perpendicular to the base here OM is the apothem. Let the side be of length‘s’ and apothem be ‘a’. The triangle OA and OB are of equal length.

Angle AOM and Angle BOM are equal and would be

$\frac{\pi}{8}$ degrees. AM would be ½ s. Consider the triangle AOM.

Area of triangle AOB = $\frac{1}{2}$($\overline{AB}$)($\overline{OM}$)$\overline{AB}$ = s

$\overline{OM}$ = $\overline{AM}$ cot ($\angle AOM$)

$\overline{OM}$ = $\overline{AM}$ cot(

$\frac{\pi}{8}$)

$\overline{OM}$ =

$\frac{s}{2}$ cot(

$\frac{\pi}{8}$)

Area of triangle AOB =

$\frac{1}{2}$(s)(

$\frac{s}{2}$cot(

$\frac{\pi}{8}$))

=

$\frac{s^2}{4}$cot(

$\frac{\pi}{8}$) we found area of on triangle.

The area of octagon can be found by multiplying area of triangle AOB with 8.

Area of Regular Octagon = 8 $\times$ ($\frac{s^2}{4}$cot($\frac{\pi}{8}$)) = 2$s^2$cot($\frac{\pi}{8}$) Area of Regular Octagon Formula = 2$s^2$cot(

$\frac{\pi}{8}$). Where ‘s’ is the side length.

To make it calculator friendly we can write it as

$\frac{2s^2}{tan(\frac{\pi}{8})}$ The area of regular octagon when circum radius instead of side length is given is found as shown below.

The area of triangle AOB when OA = OB = r is given is found by =

$\frac{1}{2}$(OA)(OB)sin($\angle AOB$)

=

$\frac{1}{2}$ (r)(r) sin(

$\frac{\pi}{4}$)

=

$\frac{1}{2}$$r^2$

$\frac{\sqrt{2}}{2}$=

$\frac{\sqrt{2}}{4}$${r^2}$

The area of the regular octagon would be 8 times

$\frac{\sqrt{2}}{4}$${r^2}$ so we get

The area of octagon formula when radius r is given is = 2$\sqrt{2^2}$

### Area of Irregular Octagon

Area of irregular octagon cannot be found by a formula. We can still
find area of irregular octagon by dividing the shape into smaller
triangles and area of each triangle can be found by the area of triangle
formulas and all triangle areas can be added up to give the area of the
irregular octagon.

Consider this irregular octagon

We choose a vertex A and from that we form different triangles keeping the vertex A as constant. We can see that from one vertex we are able to form 6 triangles.

In
this figure the triangles are AHG, ABC, ADC, ADE, AEF, AFG We need to
area of each triangle then add them up to get the area of the octagon.

### Solved Examples

**Question 1: **Find the area of a regular octagon given that its side length is 4 cm.

** Solution: **

The formula for area of regular octagon = $\frac{2s^2}{tan(\frac{\pi}{8})}$

Given s = 4 cm

Area = $\frac{2(4)^2}{tan(\frac{\pi}{8})}$ = $\frac{2(16)}{tan(\frac{\pi}{8})}$ = $\frac{32}{tan(\frac{\pi}{8})}$

Using a calculator

Area = 77.25 sq. cm.

**Question 2: **Find the area of regular octagon whose circum radius is 5cm

** Solution: **

The formula for area of regular octagon when circum radius ‘r’ is given = 2$\sqrt{2r^2}$

Here r = 5cm

Area = 2$\sqrt{2(5)^2}$ sq. cm

Area = 2$\sqrt{2(25)}$ sq. cm.

Area = 50$\sqrt{2}$ sq. cm.