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Equable Polygons

Date: 10/29/2001 at 11:33:09
From: Liz Ely
Subject: Equable polygons

I am currently doing a piece of coursework on equable shapes and I 
have to find a formula for finding equable polygons. I have noted the 
formula 4/tan(90-180/n), but how do you get to this point? Where is 
the correlation and how would you go about finding the answer to this 
question? what I mean to say is I have the question and the answer but 
no reasoning. Help, please!

Date: 10/29/2001 at 12:29:24
From: Doctor Peterson
Subject: Re: Equable polygons

Hi, Liz.

Did you find that formula by searching our archives? If so, you should 
have found several other answers that tell HOW to get this sort of 
formula. Try going to our search page   

and entering "equable" to see what we've said. You'll have to do some 
work to get it, which is why I suggest starting out with simpler cases 
to get a feel for how it works. Try finding the side of an equable 
square first.

- Doctor Peterson, The Math Forum   

Date: 10/29/2001 at 13:54:12
From: Liz Ely
Subject: Re: Equable polygons

Hi - 

Thank you for answering my question, but I have already done as you 
suggest; in fact I have nearly finished my coursework - it is just the 
way to get to the formula ( 4/tan(90-180/n)) that puzzles me. I have 
completed square - triangle - circle - hexagon - pentagon - even most 
3d shapes, but I am stuck on how to connect all the regular polygons 
with the formula and how I could find the formula myself, without 
saying that I have simply found it on the Internet.

Date: 10/29/2001 at 14:30:29
From: Doctor Peterson
Subject: Re: Equable polygons

I wish you had shown me how you tried to find this formula for the 
polygon; it sounds as if you know what to do, so you must have become 
stuck at some point in doing the algebra or trigonometry. I'm trying 
to guide you through finding the formula yourself, but I need to see 
just where you need help.

I'll assume you found the formulas for the area and perimeter of a 
regular polygon here:

   Regular Polygon Formulas - Dr. Math FAQ   

Use these formulas from that page, giving perimeter and area in terms 
of side length a and number of sides n, together with angle alpha, 
which is equal to 360/n degrees:

   P = na
   K = na^2 cot(alpha/2)/4 

Now set P = K and solve for n. That's all you have to do!

The formula on that page has a rather odd form, which the author must 
have obtained using a somewhat different starting point. You can 
simplify it to get a nicer formula that you can more easily compare 
with whatever you get. Just note that tan(90-x) = cot(x), and that 
cot(x) = 1/tan(x).

- Doctor Peterson, The Math Forum   

Date: 04/14/2002 at 17:15:08
From: Mike
Subject: Why use 4 in equable shape formula

Your equable shape formula: 


doesn't say why you used the number 4 in the formula. I understand 
the rest of the formula but I just don't understand why 4 works. Why 
not 2 or 3 ?


Date: 04/14/2002 at 17:50:41
From: Doctor Schwa
Subject: Re: Why use 4 in equable shape formula

Hi Mike,

An n-sided polygon of side length a has a perimeter of na. 
Its area is n * the area of each triangle made by connecting the 
center to two adjacent vertices.
The base of each such triangle is a.
The height is found by using tan(360/(2n)) = (a/2) / height, so
height = (a/2) / tan(180/n).

So the area is n * (1/2) * (a) * (a/2) / tan(180/n).
Setting those equal gives na = na^2 / (4 tan (180/n)), so
a = 4 * tan(180/n).

So the 4 comes from 2*2: one of the 2s is the 1/2 in the formula for 
the area of a triangle, and the other 2 is because the side of the 
polygon gets cut in half when you're making a right triangle to find 
the height.

Finally note that tan(90-x) = 1/tan(x), if you want to convert
the formula into that slightly strange form in the above answer.

- Doctor Schwa, The Math Forum
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Triangles and Other Polygons

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