## Area of Octagon:

An octagon is a polygon with eight sides; a polygon being a two-dimensional closed figure made of straight line segments with three or more sides.The word octagon comes from the greek *oktágōnon*, “eight angles”. In a regular polygon, there are 8 sides of equal length and equal internal angles – 135^{0}. An irregular octagon is one which has 8 different sides. Irregular polygons are not considered as having a centre and are not symmetric. The STOP sign used in many countries comes in the shape of an octagon. In this article, find out various methods and formulae to calculate area of octagon.

** **

There are several ways to calculate the area of octagon. Let us look at some of the methods.

** ****For regular octagon**

__#Method 1__

__#Method 1__

The octagon can be conceptualized as small eight isosceles triangles sharing a common apex. Area of one of the triangles is calculated, and then we can multiply the result by 8 to find the total area of octagon.

Consider one of the triangles. Draw a line from the midpoint of the base to the apex to form a right angle. Thus, the base of the triangle is ** a**, which is the length of the side of the polygon and OD is the height of the triangle.

Area of the octagon is given as 8 x Area of Triangle.

2 sin²θ = 1- cos 2θ

2 cos²θ = 1+ cos 2θ

Area of the octagon = 8 x Area of Triangle

Area of Octagon =

Hence,

**Area of Octagon = **

**Also,**

The area of a regular octagon of side length *a* is given by

A = 2 * cot pi / 8 * a^{2 }

A = 2 cot π 8 a 2 = 2 ( 1 + 2 ) a 2 ≃ 4.828 a 2 . {\displaystyle A=2\cot {\frac {\pi }{8}}a^{2}=2(1+{\sqrt {2}})a^{2}\simeq 4.828\,a^{2}.} In terms of the circumradius *R*, the area is

A = 4 * sin pi/4 * R^{2 }

A = 4 sin π 4 R 2 = 2 2 R 2 ≃ 2.828 R 2 . {\displaystyle A=4\sin {\frac {\pi }{4}}R^{2}=2{\sqrt {2}}R^{2}\simeq 2.828\,R^{2}.}

In terms of the apothem *r*, the area is,

A = 8 * tan pi/8 * r^{2 } where r = (1+ √2)/2 * a

**#Method 2 **

If the octagon is subdivided into non-overlapping pieces, then we get a square, four rectangles, and four isosceles right angled triangles.

So now, we calculate area of the square A_{sq }= a^{2 },

The area of the triangle is A_{tr }= ½ * x*x

Where x = √[(a^{2 }/2)] (since, for a right angled triangle, base^{2 }+ height^{2 }= square of hypotenuse = side of the octagon).

Area of rectangle A_{rect} = x * a

Thus, area of the polygon = A_{sq }+ 4 * A_{rect }+ 8 * A_{tr }square units.

Image Credits: Math Central

__#Method 3__

__#Method 3__

Image Credits: Dummies

The octagon is a square with four triangles cut off from each corner of the square.

Thus, the side of the octagon forms the hypotenuse of these triangles.

A^{2}= 2x^{2}

Thus, x = √[(a^{2 }/2)]

Therefore, the length of each side of the square is l = a+ 2x = a + 2*√[(a^{2 }/2)]

Therefore, area of the octagon is A = (l*l) – 4 * (1/2* x*x) sq. units,

where x = √[(a^{2 }/2)]

__#Method 4__

__#Method 4__

** **Consider the diagram below with radius r:

A regular octagon can be thought of as being composed of 4 “kite” shaped areas.

The area of a “kite” with diagonals d and w is:

Area kite=d⋅w / 2.

(This is fairly easy to prove if it isn’t a formula you already know).

Consider the “kite” PQCW in the diagram above.

∠QCW=π2 and |QC|=|WC|=r

XXX⇒|QW|=√2r (Pythagorean)

Therefore (since |PC|=r)

AreaPQCW=|PC|⋅|QW|2=r⋅√2r^{2}=√^{2r2 }/ 2

The octagon is composed of 4 such kites, so

Area of octagon=2√2r^{2}

where r is the radius of the octagon

**Irregular octagons**

Break the octagon into triangles, and then add the areas of individual triangles.

Area of individual triangles can be found by Heron’s formula, which says:

If the sides of the triangle are a, b, c, and the semi perimeter is s (s = (a+b+c)/2), then the area is

**Problems**

**Question 1: **Find the area of an octagon of side 10 cm?

**Solution: **Given,

Side of an octagon = a = 10 cm

Area of an octagon

= 2 ×

a^{2} × (1 + √2)

= 2 × 10^{2} × (1 + √2) cm^{2}

= 2 × 100× (1 + 1.414) cm^{2}

= 482.8 cm^{2 }

^{ }

**Question 2:** Perimeter of an octagonal stop sign board is 48 cm. Find the area of the signboard.

**Solution**: Given,

Perimeter of the stop sign board = 48 cm

Perimeter of an Octagon = 8a

48 cm = 8a

a = 48/8 = 6 cm

Area of an Octagon =

Area of the stop sign board = 2* 6*6 *2.414 = 173.81 cm sq.

**Question 3.** Find the area of a regular octagon with 0.8 cm area

**Solution**: Subdividing it into a square, four rectangles and four triangles,

Area of the square = 0.8 * 0.8 = 0.64

Area of the triangle and rectangle depends on the value of square. Thus, we need to find out x.

X = √ [(a*a)/2] = √(0.64/2) = 0.57

Now, area of triangle = ½ * x * x = 0.16

Area of rectangle = x*a = 0.57 * 0.8 = 0.46

Thus, area of octagon = 0.64 + 4*0.16 + 4 * 0.46 = 0.64 + 1.824 + 0.64 = 3.104 cm. sq.

** **

**Question 4**. Find the area of a regular octagon whose side is 50 cm.

**Solution: **

Given that, side s = 50 cm.

We know that the area of a regular octagon = side2×(2+2(2–√))

= 502×(2+2(1.414))

= 2500×(2+2.828)

= 2500×(4.828)

= 12070 cm^{2}

So, the above demonstration described several methods to calculate the area of an octagon. Also check out our article on Area of Pentagon here.