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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.

Thanks,
Terry
```

```
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

and

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
http://mathforum.org/dr.math/
```

```
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.

Regards,
Terry.
```

```
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.

- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/
```

```
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
perfectly!

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
http://mathforum.org/dr.math/
```

```
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
http://mathforum.org/dr.math/
```

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

Thanks,

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.
Regards,
Terry.
```
Associated Topics:
College Polyhedra
High School Polyhedra
High School Polynomials

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