# Diagonal of a Cuboid

• October 17th 2011, 07:32 AM
BobtheBob
Diagonal of a Cuboid
The total area of all the faces of a cuboid is 22 cm2, and the total length of all its edges is 24 cm. Find the length of any one of its internal diagonals.
I can't seem to get this, so any help appreciated
• October 17th 2011, 07:53 AM
Plato
Re: Diagonal of a Cuboid
Quote:

Originally Posted by BobtheBob
The total area of all the faces of a cuboid is 22 cm2, and the total length of all its edges is 24 cm. Find the length of any one of its internal diagonals.

Suppose that $a,~b,~\&~c$ are the lengths of the sides.
The length of the diagonal is $\sqrt{a^2+b^2+c^2}$.
Hint: what is $(a+b+c)^2~?$
• October 17th 2011, 08:42 AM
BobtheBob
Re: Diagonal of a Cuboid
Quote:

Originally Posted by Plato
Suppose that $a,~b,~\&~c$ are the lengths of the sides.
The length of the diagonal is $\sqrt{a^2+b^2+c^2}$.
Hint: what is $(a+b+c)^2~?$

Ok thanks for the reply. I've already managed to get these equations from the question:

22=2(ab+ac+bc)

24=4(a+b+c)

d= http://latex.codecogs.com/png.latex?...b%5E2+c%5E2%7D.

but I can't manage to work out how to get the lengths?
• October 17th 2011, 08:55 AM
Plato
Re: Diagonal of a Cuboid
Quote:

Originally Posted by BobtheBob
Ok thanks for the reply. I've already managed to get these equations from the question:
22=2(ab+ac+bc)
24=4(a+b+c)
d= http://latex.codecogs.com/png.latex?...b%5E2+c%5E2%7D.
but I can't manage to work out how to get the lengths?

We do not need the lengths.
All we need to know is $a^2+b^2+c^2=~?$
Note that
$(a+b+c)^2=a^2+b^2+c^2+2ac+2ac+2bc$

BTW: You can use LaTeX directly here.
[TEX]\sqrt{a^2+b^2+c^2}[/TEX] gives $\sqrt{a^2+b^2+c^2}$
• October 17th 2011, 09:07 AM
BobtheBob
Re: Diagonal of a Cuboid
Quote:

Originally Posted by Plato
We do not need the lengths.
All we need to know is $a^2+b^2+c^2=~?$
Note that
$(a+b+c)^2=a^2+b^2+c^2+2ac+2ac+2bc$

BTW: You can use LaTeX directly here.
[TEX]\sqrt{a^2+b^2+c^2}[/TEX] gives $\sqrt{a^2+b^2+c^2}$

Ok, thanks for the tip.

Where have you got $(a+b+c)^2$ from? I'm really sorry but I think I'm being a bit slow today. (Speechless)

I know that $a^2+b^2+c^2=Diagonal$ because you get that from pythagorus' theorum, but where do you get $(a+b+c)^2$ and $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$ from?

Thanks again and sorry that I'm being a bit slow today.
• October 17th 2011, 09:16 AM
Plato
Re: Diagonal of a Cuboid
Quote:

Originally Posted by BobtheBob
$\color{red}22=2(ab+ac+bc)$
$\color{blue}24=4(a+b+c)$

Look at what you already know.
$\color{red}{2ab+2ac+2bc=22}~\&~\color{blue}{a+b+c= 6}$

So what is $a^2+b^2+c^2~?$
• October 17th 2011, 09:22 AM
BobtheBob
Re: Diagonal of a Cuboid
Oh right so $a^2+b^2+c^2=36$?!
• October 17th 2011, 09:25 AM
Plato
Re: Diagonal of a Cuboid
Quote:

Originally Posted by BobtheBob
Oh right so $a^2+b^2+c^2=36$?!

Absolutely not.
$(a+b+c)^2=36$
• October 17th 2011, 09:59 AM
BobtheBob
Re: Diagonal of a Cuboid
Quote:

Originally Posted by Plato
Absolutely not.
$(a+b+c)^2=36$

Oh right of course!

So

$2ab+2ac+2bc=22$

$4a+4b+4c=24$
Therefore $a+b+c=6$

Then, $(a+b+c)^2=36$

$a^2+b^2+c^2+2ab+2ac+2bc=36$
• October 17th 2011, 10:07 AM
BobtheBob
Re: Diagonal of a Cuboid
Quote:

Originally Posted by Plato
Absolutely not.
$(a+b+c)^2=36$

Oh right of course!

So

$2ab+2ac+2bc=22$

$4a+4b+4c=24$
Therefore $a+b+c=6$

Then, $(a+b+c)^2=36$

$a^2+b^2+c^2+2ab+2ac+2bc=36$

So sub in $22=2ab+2ac+2bc$

$36=a^2+b^2+c^2 +22$

$a^2+b^2+c^2=14$

Therefore,
$\sqrt{a^2+b^2+c^2}=\sqrt{14}$

So the diagonal is $\sqrt{14}$ as $d=\sqrt{a^2+b^2+c^2}$

Thanks so much for the help! It's really obvious when you know. (This is right though?) (Happy)