## Introduction: Drawing a Pentagon

This is a companion to the Dodecahedron Calendar project. Here I demonstrate how to draw a perfect pentagon. You will need a ruler and a compass.

## Step 1:

First draw the horizontal and vertical lines.

## Step 2:

Then draw the circle centered on the crosshairs. Do not fold the compass after the circle is drawn. The side of the pentagon will be 1.176 times the radius.

## Step 3:

Without adjusting the compass, place the point of the compass on the circle where it crosses the horizontal line. Now draw arcs on the previous circle above and below and connect those points.

## Step 4:

Now center the compass on the crosshair made from the bisector and draw an arc from the top of the circle down to the horizontal line. BTW, you've just drawn a golden ratio.

## Step 5:

At the end of this step, do not close the compass. You will need that distance to make four more arcs. Putting the point of the compass at the top of the circle draw an arc from where the last arc intersected the horizontal line out to the circle.

## Step 6:

Now move around the circle using each arc as the center of the next arc.

## Step 7:

And finally, draw lines from each intersection to form a pentagon.

To be honest, I've never gotten this to work the first time. There are just too many places to make a few thousandths of an inch mistake and they all add up. What I have to do is place the center of the compass at the arc intersection just above the horizontal line on the right side, then adjust the compass in or out by one fifth of the distance I'm off from the top of the circle. Then start over from the top of the circle and work my way around again. Even though this sounds like proof that the process is wrong I've done the math and it should work if not for human error.

## 3 People Made This Project!

- Legendgamer722 made it!
- Jack A Lopez made it!
- Jack A Lopez made it!

## 21 Comments

Question 4 weeks ago

Draw a regular pentagon with size 60mm

Question 3 months ago on Step 5

Sir how did u get the arc ??

Question 8 months ago on Introduction

in the first step, wherefrom has this 1.176 come?

Answer 5 months ago

I started by deciding how long a side should be. Then using the 1.176 I found the radius that gave me that length side on the pentagon.

10 months ago

I’ve tried to do this multiple times, but am never able to get past step 4. The arc never intersects with the central vertical line. The arcs that I make after that step are always too short.

Reply 7 months ago

I ran into the same trouble as you. I thought at first the algorithm was off by a little bit, but then I realized I didn't understand the instructions fully. I wrote a detailed explanation in response to another user's question ("balwantmay"), if you are interested.

Question 8 months ago on Step 2

Sorry. I mentioned step 1, it is on step 2. I could not understand the term'`1.176 times the radius. I could not understand the rationale of 1.176 . Kindly elucidate. regards bs[age 81]

Answer 7 months ago

1.176 can be computed using the sine function, i.e., 2*sin(36). You can see why this is the right value by dividing your circle into 5 congruent triangles, each with an angle at the center of the circle equal to 72 degrees (i.e., 360/5). Cut this angle in half to form a right triangle (shown in light blue in the diagram), so that you can make use of the sine function to compute half the length of the side of the pentagon (where here for simplicity the radius of the pentagon is assumed to be 1, i.e., a unit circle).

The approach given in this Instructable to draw a pentagon is correct, but a lot of people get off track in step 5. Perhaps not sufficiently clear is that you have to set your compass a second time, or else the side length will be too short. The second diagram illustrates that the side length generated

if you don't set the compass the second timeis actually one half the root of 5, which is approximately 1.118, or about 5% off from 1.176.BTW, the logic behind the calculations in this second diagram is that the chord formed by the radius is subtended by a 60 degree angle (you can see this by inscribing a regular hexagon in the circle), from which you get the 60 degree angle on the right, from which we can form a 30-60-90 triangle, which with a little work (not shown) involving the pythagorean theorem leads to the exact length generated using this compass and straightedge method.

The image by Jack Lopez (under "I Made it!") shows how the math ties out. Thanks Jack for that effort, as it helped me see why I had originally gotten into trouble using this method myself (i.e., I didn't set the compass the second time).

1 year ago

I adjust my compass to have a span equal to the hypotenuse of the triangle created by the bisection of the right-side radius, use that to scribe my arc to create the golden ratio segment of the left-side radius, then use the same span of my compass to scribe the second arc on the circumference of the circle, using each arc as the center of the next, but when I get to my last one the distance from the top point to the left point isn't the same as the distance from the top point to the one on the right! And it should be, correct? And yet, when I connect all my points with straight lines, all the intersections have congruent angles... all the lines are the same length, all the points are the same distance from each other... look how regular my result looks!

(noise of mental short circuit)

It's driving me nuts why this isn't so! Also, I want to know WHY this is how it's done... what's the reasoning behind using the hypotenuse to create a golden ratio?

1 year ago

Thank u sir I appreciate it

Question 2 years ago on Step 2

The distance

Answer 2 years ago

I am not sure what you're asking here, but yesterday I did this same construction, using pencil-paper-compass-and-straightedge, and I uploaded a picture of this to the Share/I-made-it section of this instructable.

The picture has the various points labeled, (A,B,C,D, etc.) and I wrote a comment to go with it that mentions the length of some of these various line segments.

By the way, if you want to see how the pros do this, the pages at Wolfram,

http://mathworld.wolfram.com/Pentagon.html

or Wikipedia,

https://en.wikipedia.org/wiki/Pentagon

might have the answers you are seeking.

5 years ago

You didn't mentioned in step 3 the extra line comes from where. ??

6 years ago

You could also use the following formula wich does not use the sin function.

t=r*sqrt(1+{(sqrt(1.25)-.5)^2})

8 years ago on Introduction

http://ru.wikipedia.org/wiki/%D0%9F%D1%80%D0%B0%D0%B2%D0%B8%D0%BB%D1%8C%D0%BD%D1%8B%D0%B9_%D0%BF%D1%8F%D1%82%D0%B8%D1%83%D0%B3%D0%BE%D0%BB%D1%8C%D0%BD%D0%B8%D0%BA

Reply 8 years ago on Introduction

Yes, you are right.

Reply 6 years ago on Introduction

Estoy luchando para dibujar un pentágono en AutoCAD, dada la longitud del lado. Por supuesto, todo lo que se requiere es el radio del círculo dentro del cual el pentágono es conocido:

Donde R es el radio y t es la longitud del lado.

Reply 6 years ago on Introduction

Bill, I tried to understand your formula, but without success. Maybe during the day I could to think something.

Reply 6 years ago on Introduction

In AutoCAD, one must inscribe a polygon inside a circle of a certain radius. For a pentagon, I know the length of a side only, do not know radius. So, using the formula above I can calculate the radius if I know the length of a side. T=1.175*R; also R=0.851*T.

En AutoCAD, hay que inscribir un polígono dentro de un círculo de un radio determinado. Para un pentágono, sé la longitud de sólo un lado, no sé radio. Así, el uso de la fórmula I anterior puede calcular el radio si conozco la longitud de un lado. T = 1.175 * R; también R = 0,851 * T.

Reply 6 years ago on Introduction

Very interesting, Bill, your deduction of the relation between R and T. Many persons would need to know it.