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Volume of a Rhombicuboctahedron


Date: 01/19/2001 at 01:47:06
From: Nicoal
Subject: Volume of rhombicuboctahedron

Dear Dr. Math,

I need the formula for finding the volume of a polyhedron called a 
rhombicuboctahedron.

Thank you
Nicoal


Date: 01/19/2001 at 13:09:15
From: Doctor Rob
Subject: Re: Volume of rhombicuboctahedron

Thanks for writing to Ask Dr. Math, Nicoal.

For a picture, see this page from MathConsult - Dr. R. Maeder:

  10: rhombicuboctahedron
  http://www.mathconsult.ch/showroom/unipoly/10.html   

I figured out the volume in the following way:

Let the length of each edge of the polyhedron be s. From its center, 
connect line segments to each of its 24 vertices. Each edge of the 
polyhedron and two of these lines to its center form a triangle. Each 
face of the polyhedron and three or four of these triangles (along 
with their interiors) form a pyramid with either a square base (18 of 
these) or equilateral triangular base (8 of these).

Now the base of each square-based pyramid has area s^2 and its height 
is s*(1+sqrt[2])/2 (found by drawing an octagonal cross-section of the 
polyhedron). Now the volume of any pyramidal region is 1/3 its base 
times its height, so the volume of each of these is:

     V = s^3*(1+sqrt[2])/6

The base of each triangular pyramid has area s^2*sqrt(3)/4, and its 
height is s*(3+sqrt[2])/(2*sqrt[3]) (found by drawing a cross-section 
at a 45-degree angle to the first one). Then the volume of each of 
these is:

     V = s^3*(3+sqrt[2])/24

Thus the total volume of the rhombicuboctohedral region is:

     V = 18*s^3*(1+sqrt[2])/6 + 8*s^3*(3+sqrt[2])/24
       = 2*(6+5*sqrt[2])*s^3/3

The radius of its circumscribed sphere turns out to be:

     R = s*sqrt(5+2*sqrt[2])/2

Its surface area is:

     S = 18*(s^2) + 8*(s^2*sqrt[3]/4)
       = 2*s^2*(9+sqrt[3])

The largest sphere that can be inscribed has radius r = s. This is 
tangent to all the square faces at their centers, but does not 
intersect any of the triangular faces, which are 

     s*(3+sqrt[2])/(2*sqrt[3]) > 1.247*s > r

away from the center.

I believe I have done the above computations correctly, but you should 
redo them to check me. They seem to check with the following site:

   Platonic And Archimedean Solids - Bruce A. Rawles
   http://www.intent.com/sg/polyhedra.html   

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Polyhedra

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