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.

You didn't mentioned in step 3 the extra line comes from where. ??
You could also use the following formula wich does not use the sin function.<br><br>t=r*sqrt(1+{(sqrt(1.25)-.5)^2})
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
Yes, you are right.
<p>Estoy luchando para dibujar un pent&aacute;gono en AutoCAD, dada la longitud del lado. Por supuesto, todo lo que se requiere es el radio del c&iacute;rculo dentro del cual el pent&aacute;gono es conocido: <br></p><p><img src="http://upload.wikimedia.org/math/9/1/3/9137244dfca50cd2740593db9ada852d.png"></p><p><br>Donde R es el radio y t es la longitud del lado.</p>
<p>Bill, I tried to understand your formula, but without success. Maybe during the day I could to think something. </p>