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How To Find The Area Of A Regular Octagon

A polygon is a ii-dimensional (2-D) airtight figure fabricated up of straight line segments. In geometry, the octagon is a polygon with 8 sides. If the lengths of all the sides and the measurement of all the angles are equal, the octagon is called a regular octagon. In other words, the sides of a regular octagon are congruent. Each of the interior angle and the exterior angle measure 135° and 45° respectively, in a regular octagon. There is a predefined prepare of formulas for the calculation of perimeter, and expanse of a regular octagon which is collectively called as octagon formula. For an octagon with the length of its border equally "a", the formulas are listed below.

Octagon Formula

Too, cheque: Octagon

Octagon Formulas

Formulas for Octagon
Area of an Octagon 2a2(1+√2)
Perimeter of an Octagon 8a

Octagon formula helps united states to compute the area and perimeter of octagonal objects.

Derivation of Octagon Formulas:

Consider a regular octagon with each side "a"units.

Formula for Expanse of an Octagon:

Area of an octagon is divers as the region occupied inside the boundary of an octagon.

In order to calculate the area of an octagon, we divide it into small eight isosceles triangles. Calculate the expanse of i of the triangles and and then we can multiply by 8 to observe the full area of the polygon.

Octagon Area Formula

Take i of the triangles and draw a line from the apex to the midpoint of the base to form a correct bending. The base of the triangle is a , the side length of the polygon and OD is the height of the triangle.

Area of an Octagon Using Triangles

Area of the octagon is given as eight 10 Area of Triangle.

2 sin²θ = 1- cos 2θ

2 cos²θ = 1+ cos 2θ

\(\brainstorm{assortment}{fifty}tan^{ii}\theta = \frac{1-cos2\theta}{1+cos2\theta}\\ tan^{2}(\frac{45}{two})=\frac{1-cos45}{1+cos45}\\ tan^{2}(\frac{45}{2})=\frac{1-\frac{i}{\sqrt{2}}}{ane+\frac{1}{\sqrt{2}}}\\ tan^{2}(\frac{45}{2})=\frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{(\sqrt{2}-1)^{ii}}{1}\\ tan(\frac{45}{2})=\sqrt{ii}-1\\ \frac{BD}{OD}=\sqrt{2}-1\\ OD=\frac{a/2}{\sqrt{ii}-ane}=\frac{a}{two}(1+\sqrt{2})\end{array} \)

Area of ∆ AOB =

\(\begin{array}{50}\frac{ane}{2}\times AB\times OD\finish{assortment} \)

=

\(\brainstorm{array}{l}\frac{1}{ii}\times a\times \frac{a}{two}(1+\sqrt{2})\cease{array} \)

=

\(\brainstorm{array}{l}\frac{a^{2}}{four}(1+\sqrt{2})\end{array} \)

Area of the octagon = 8 x Area of Triangle

Area of Octagon =

\(\begin{array}{l}eight\times \frac{a^{2}}{4}(1+\sqrt{ii})\end{array} \)

Area of an Octagon =

\(\begin{array}{l}2a^{ii}(1+\sqrt{2})\end{array} \)

Formula for Perimeter of an Octagon:

Perimeter of an octagon is defined as the length of the boundary of the octagon. And so perimeter will be the sum of the length of all sides. The formula for perimeter of an octagon is given by:

Perimeter = length of 8 sides

So, the perimeter of an Octagon = 8a

Properties of a Regular Octagon:

  • It has eight sides and eight angles.
  • Lengths of all the sides and the measurement of all the angles are equal.
  • The full number of diagonals in a regular octagon is xx.
  • The sum of all interior angles is equal to 1080 degrees, where each interior angle measures 135 degrees.
  • The sum of all exterior angles is equal to 360 degrees, where each exterior angle measures 45 degrees.

Solved examples Using Octagon Formula:

Question 1: Calculate the area and perimeter of a regular octagon whose side is 2.3 cm.

Solution: Given, side of the octagon = 2.iii cm

Area of an Octagon =

\(\brainstorm{array}{l}2a^{2}(1+\sqrt{2})\terminate{assortment} \)

Area of an Octagon =

\(\begin{assortment}{fifty}2\times two.iii^{2}(1+\sqrt{ii})=25.54\;cm^{ii}\end{assortment} \)

Perimeter of the octagon= 8a= viii × 2.3 = 18.4 cm

Question 2:Perimeter of an octagonal stop signboard is 32 cm. Find the area of the signboard.

Solution: Given,

Perimeter of the cease sign board = 32 cm

Perimeter of an Octagon = 8a

32 cm = 8a

a = 32/8 = 4 cm

Expanse of an Octagon =

\(\begin{assortment}{l}2a^{2}(1+\sqrt{ii})\stop{array} \)

Area of the stop sign board =

\(\brainstorm{assortment}{l}two\times 4^{2}(i+\sqrt{2})=77.248\;cm^{2}\terminate{array} \)

To solve more bug on the topic, download BYJU'Southward – the Learning App.

Source: https://byjus.com/octagon-formula/

Posted by: hendrixdesten.blogspot.com

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