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Find the Edge Lengths of a Cuboid

Date: 05/13/2003 at 09:14:07
From: Terry
Subject: Edge lengths of  a cuboid

Hi Doctor,

I can ascertain the following characteristics of a cuboid;  length of 
edge 1, length of edge 2, length of edge 3, diagonal, area and volume 
when I know any two of the edge lengths and any one of the diagonal, 
area or volume.  However, I am trying to discover the lengths of the 
edges when only the diagonal, area, and volume are known. Could you 
point me in the right direction please?

I know the combined length of the edges = the cube root of the volume 
and that the square of the diagonal is equal to the sum of the squared 
edge lengths.


Date: 05/13/2003 at 16:19:01
From: Doctor Douglas
Subject: Re: Edge lengths of a cuboid

Hi Terry,

Thanks for writing to the Math Forum.

If you know the Volume V of the cuboid and the area A of one of the
faces, then 

  p = V/A 

is the length of the cuboid perpendicular to the face A.

It's not clear what you mean by "diagonal," but let's say that it is 
the "body diagonal" that stretches from one corner to the opposite 
corner, going through the cuboid's interior.

Then if this diagonal is D, then

  d = sqrt(D^2 - p^2)  is the length of the diagonal of the face 
                       whose area we know.

By the Pythagorean theorem, if q and r are the edge lengths of
this face, then

  q^2 + r^2 = d^2


  p * q * r = V

Because we already know p (and d), these last two equations are
two equations in two unknowns (in q and r), which we can solve
to obtain the length of the other dimensions.
- Doctor Douglas, The Math Forum 

Date: 05/14/2003 at 06:27:41
From: Terry
Subject: Edge lengths of  a cuboid

Hi Doctor Douglas,

Thanks for responding.  

Sorry I wasn't clear on the diagonal, what I meant was the space 
diagonal.  If you are looking down on an empty shoebox, the space 
diagonal would run from the front top left vertex to the back bottom 
right vertex.

I understand what you have said, but I am trying to ascertain all 
three edge lengths when I do not know any of them. The only 
information available is the total surface area, the volume, and the 
space diagonal. Unfortunately, your solution assumes knowledge of the 
surface area of one of the faces - which is not known. 

I would also hazard a guess that if these are the only measurements 
available then more than one solution might be possible on occasion.


Date: 05/14/2003 at 18:40:15
From: Doctor Douglas
Subject: Re: Edge lengths of  a cuboid

Hi Terry,

Thanks for writing back.  The information we are given is

  D^2 = a^2 + b^2 + c^2                D = body or space diagonal

  V = abc                              V = total volume

  A = 2ab + 2bc + 2ac                  A = total surface area

D, V, and A are given and we are looking for a triplet (a,b,c) that 
satisfies the three equations above. We do not require knowing a, b, 
and c individually, because they may be permuted, and this leads to 
essentially the same cuboid.

This looks like three equations in three unknowns, and so it should be 
possible to determine the solution. There may be a clever way to the 
solution, but I don't know it. One could simply eliminate variables 
one by one, and end up with an equation that may have to be solved 
numerically for the final undetermined variable.

I'll think about this some more.

- Doctor Douglas, The Math Forum 

Date: 05/21/2003 at 14:50:51
From: Doctor Schwa
Subject: Re: Edge lengths of  a cuboid

Hi Terry,

Symmetric polynomials are one of my great loves, and this problem fits 

There are two clever tricks you can use in this type of situation. You 
found one, already: you can solve for (a+b+c) or any other symmetric 
combination of a,b,c that you want. But you want to solve for a, b, c 
separately: that's hard. Among other problems, you have no way of 
knowing which of the three sides should be called a, b, or c: there 
are at least those 6 solutions, from rearranging the three variables.

The second clever trick is to let x = an unknown side length. Then you 
can take a look at the polynomial (x-a)(x-b)(x-c) = 0, which will have 
the three side lengths you want as its three solutions. Then it's a 
symmetric combination of a, b, c again:

   x^3 - (a+b+c)x^2 + (ab + ac + bc)x + abc = 0

Now that you know the values of a+b+c, and ab+ac+bc, and abc, you can 
solve this cubic (by using the cubic formula, if you must; it's in our 
archives/FAQ) to find the three values of x which will be the three 
side lengths.

Hope that helps,

- Doctor Schwa, The Math Forum 

Date: 05/21/2003 at 17:41:20
From: Doctor Douglas
Subject: Re: Edge lengths of  a cuboid

Hi again, Terry,

Dr. Schwa's solution is very beautiful, because it dissolves all of 
the ugly algebra so efficiently:

   x^3 - (a+b+c)x^2 + (ab+bc+ac)x - abc = 0      [Note:  -abc]

which, in terms of your original data, is

   x^3 - sqrt(D^2 + A)x^2 + (A/2)x - V = 0

Since your data come as real positive numbers D, A, and V, the
coefficients in this polynomial alternate in sign, and it is 
reassuring to check that Descartes Rule of Signs permits three 
possible solutions for x, since there are three sign changes.

Now it is just a matter of finding the roots of this polynomial,
and you can resort to whatever methods you wish. 

- Doctor Douglas, The Math Forum 

Date: 05/23/2003 at 10:04:42
From: Terry
Subject: Thank you (Edge lengths of  a cuboid)


I have spent countless hours trying to get a grip on that problem.  I 
shall now trot off to Dr Math's FAQ and re-teach myself how to solve 
cubic equations - my school days were many moons ago!

Thanks again.
Associated Topics:
College Polyhedra
High School Polyhedra
High School Polynomials

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