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

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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.

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

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Originally Posted by Plato

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

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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?

Quote:

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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?

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

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

Quote:

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Originally Posted by Plato

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

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

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Ok, thanks for the tip.

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

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

Thanks again and sorry that I'm being a bit slow today.

Quote:

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Originally Posted by BobtheBob

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

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

So what is $\displaystyle a^2+b^2+c^2~?$

Oh right so $\displaystyle a^2+b^2+c^2=36$?!

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Originally Posted by BobtheBob

Oh right so $\displaystyle a^2+b^2+c^2=36$?!

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Absolutely not.
$\displaystyle (a+b+c)^2=36$

Quote:

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Originally Posted by Plato

Absolutely not.
$\displaystyle (a+b+c)^2=36$

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Oh right of course!

So

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

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


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

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

Quote:

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Originally Posted by Plato

Absolutely not.
$\displaystyle (a+b+c)^2=36$

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Oh right of course!

So

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

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


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

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

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

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

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

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

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


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