# Octagon

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 Sub Topics A polygon is closed shape with equal number of angles and sides. The sides are called edges and the two sides intersect to form vertices. A polygon with n sides would have n vertices and n angles. The Greek word “octa” means 8 and “gon” means angles thus a polygon with eight angles is called an Octagon. The octagon also has 8 sides and can be classified regular; irregular based on the sides lengths. It can also be classified as concave or convex type based on the internal angles.Some commonly seen octagon in real lifeAn open umbrella is like an octagon.The stop sign we see is also in shape of an octagon.

## Octagon Definition

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A figure with eight sides and 8 angles is called an Octagon. Another way of defining this would be a polygon with 8 sides.

## Regular Octagon

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An Octagon whose all the eight sides are equal is known as a regular Octagon. In regular Octagon all the interior angles would be the same as well.

## Irregular Octagon

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An Octagon whose sides of different lengths is called irregular octagon.

The figure below is an example of irregular Octagon. See how the sides are not equal.

## Convex Octagon

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An octagon is convex if its every interior angle lies between 0 and 180.  We can also draw a line between two vertices it would remain within the boundary of the polygon. All the figures we have seen so far are of convex octagons.

## Concave Octagon

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A concave octagon is a polygon when one or more interior angles can be more than 180°. If we draw line through any two of its vertices some part of the lines might lie outside the area enclosed by the octagon..

Given below are the figures of concave Octagon.

## Area of Regular Octagon

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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.

## Interior Angles of Octagon

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Octagon has 8 interior angles. Regular octagon has all its interior angles equal.

The sum of all the interior angles of any octagon is 1080° this can be easily found by using the formula which we use to find the sum of interior angles of regular polygon using n = 8.
The sum of interior angles of a Polygon = (n-2)*180°

The sum of angles a regular Octagon = (8-2)*180°= 6 * 180° = 1080°
The regular polygon has all its interior angles equal therefore each interior angle can be found by dividing the sum of all the angles by 8.

Each interior angle of regular octagon = $\frac{1080^o}{8}$

Each interior angle of regular octagon =  135°

## Exterior Angle of an Octagon

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An exterior angle of a polygon makes a linear pair with interior angle. i.e. the sum of the interior and exterior angles will be 180°.

From the figure you can see that we can have two angles at each vertex. But we choose only one and in this figure we have selected the clockwise arrangement of exterior angles.
The sum of exterior angles of any convex polygon would always be 360°Since an octagon is a polygon and for convex octagon this would hold good as well.

For irregular octagon each exterior angle would be of different measurement. For regular Octagon each exterior angle is found by dividing $\frac{360^o}{8}$ we get 45°.

The measure of each exterior angle of a regular Octagon is 45°