## What Are The Octagon Formulas?

The octagon is an 8-sided polygon. An octagon is referred to as a regular octagon if all of its sides have equal lengths and angles are of equal measures. A regular octagon has each interior angle measuring $135^{\circ}$ and each exterior angle measures $45^{\circ}$.

We will discuss octagon formulas such as the area of an octagon, perimeter of an octagon.

#### Begin here

## The Formula for the Area of an Octagon

Area of an octagon is the region bounded within the boundaries of an octagon. The **area of an octagon **$= 2a^{2}(\sqrt{2} + 1)$

where a represents the side of the octagon.

## Derivation

To find the area of an octagon, it is divided into 8 equal isosceles triangles. The entire area of the polygon can be determined by multiplying the area of one triangle by 8.** **

One isosceles triangle is shown below where $OA = OB$.

Here, OD is the height of the triangle, and AB is the base of the triangle. Let OD bisects the angle AOB and the side AB.

Here, base $= AB = a$

**Area of octagon** $= 8 \times\; Area\; of\; one\; isosceles\; triangle$

By using the identities, we have

$tan^{2}(\frac{\theta}{2}) = \frac{1 \;-\; cos\; \theta}{1 \;+\; cos\;\theta} = \frac{1\;-\; cos\; 45^{\circ}}{1\;+ \;cos\; 45^{\circ}} = \frac{1-(1\sqrt{2})}{1+(1\sqrt{2})} = \frac{\sqrt{2}\;-\;1}{\sqrt{2}\;+\;1} = (\sqrt{2}\;-\;1)^{2}$

$tan\; (\frac{45^{\circ}}{2}) = \frac{BD}{OD} = \sqrt{2}\;-\;1$

Thus, $OD = \frac{BD}{\sqrt{2}\;-\;1} = \frac{a/2}{\sqrt{2}\;-\;1} = \frac{a(\sqrt{2}\;+\;1)}{2}$

Area of $\Delta AOB\; = 12 \times AB \times OD$

$= \frac{1}{2} \times a \times \frac{a}{2}(\sqrt{2}\;+\;1)$

$= \frac{1}{4}a^{2}\;(\sqrt{2}\;+\;1)$

**Area of octagon **$= 8 \times$ **Area of one isosceles triangle**

Area of octagon $= 8 \times \frac{1}{4}a^{2}\;(\sqrt{2}\;+\;1)$

Area of octagon $= 2a^{2}\;(\sqrt{2}\;+\;1)$

## Formula of the Perimeter of Octagon

**The length of an octagon’s boundary is referred to as its perimeter**. Therefore, the perimeter will equal the total length of all sides. The formula for the perimeter of a regular octagon having 8 sides each measuring “a” units is given by

**So, the perimeter of a regular octagon = 8a units**

## Octagon Formulas

Area of a regular octagon $= 2s^{2}\;(1\;+\;\sqrt{2})$

Perimeter of a regular octagon $= 8s$

## Properties of a Regular Octagon

- There are 8 sides and 8 interior angles.
- A regular octagon has a total of 20 diagonals.
- Each inside angle measures $135^{\circ}$. The sum of interior angles is $1080^{\circ}$.
- Each exterior angle measures $45^{\circ}$, so the sum of all the exterior angles is $360^{\circ}$.

## Conclusion

In this article, we learned about octagon, octagonal formulas, derivation of octagon formula, and properties of a regular octagon. Let’s solve a few examples now.

## Solved Examples based on Octagon Formulas

**1. If the length of an octagon’s side is 14 inches, calculate its area.**

**Solution:**

The side’s length, “s,” is 14 inches.

Using the octagon’s surface area formula,

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

$A = 2 \times14^{2}\;(\sqrt{2}\;+\;1)$

$A = 946.37$ square inches

As a result, the octagon has a surface area of 946.37 square inches.

**2. The area of an octagon is 25.54 square units. Find the length of the side.**

**Solution:**

The octagon’s area is 25.54 square units.

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

$25.54 = 2 \times s^{2}\;(1\;+\;\sqrt{2})$

$s = 2.3$ units

The octagon’s side length is 2.3 units.

**3. If a normal octagon’s side is 5 units long, calculate its area and perimeter.**

**Solution:**

An ordinary octagon has equal-length sides.

The side length in this case is 5 units, or “s.”

Area of a regular octagon $= 2s^{2}\;(1 \;+\; \sqrt{2}) = 2 (5)^{2} (1 \;+\; \sqrt{2}) = 50 (1 \;+\; 2) = 120.71$ square units

Perimeter of a regular octagon $= 8s = 40$ units

## Practice Problems on Octagon Formulas

## Octagon Formula For Area and Perimeter With Derivation

### Determine the perimeter of a regular octagon with side 2 inches.

Solution: Side of the octagon $= 2$ inches

The perimeter of a regular octagon, with side length ‘a’, is given by the following formula,

The perimeter $= 8a = 8 \times 2 = 16$ inches

### An octagon has a 40-inch perimeter. Find the area.

Octagon’s perimeter $= 8a$

32 inches $= 8a$

$a = \frac{40}{8} = 5$ inches

Octagon Area $= 2 \;a^{2}\; (1\;+\;\sqrt{2}) = 2 \times 5^{2}\; (1\;+\;\sqrt{2})\;\simeq 120\; inches^{2}$

### Each interior angle of a regular octagon measures _____ degrees.

Each interior angle of a regular octagon measures 135 degrees.

## Frequently Asked Questions about Octagon Formulas

**What are the exterior angles of a regular octagon?**

The exterior angles are the angles that form between a side and the extension of the side adjacent to it. A regular octagon’s interior angles are all $135^{\circ}$, so the exterior angle is measured as the angle supplement to it, $180^{\circ} \;-\; 135^{\circ} = 45^{\circ}$.

**How many diagonals does an octagon have?**

The number of diagonals of a polygon of n sides is given by $\frac{n(n-3)}{2}$, substituting the value $n = 8$, we have 20 diagonals.

**Can we derive a formula for finding the area of an irregular octagon?**

Since an irregular polygon (with more than 3 sides, heron’s formula for a scalene triangle) has no fixed dimensions, we cannot find a direct formula for its area. We can divide the octagon into separate triangles and find the sum of the areas of those triangles.