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
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.
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.
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?)