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Thread: Diagonal of a Cuboid

  1. #1
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    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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  2. #2
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by BobtheBob View Post
    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 $\displaystyle a,~b,~\&~c$ 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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  3. #3
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by Plato View Post
    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 $\displaystyle (a+b+c)^2~?$
    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= .

    but I can't manage to work out how to get the lengths?
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  4. #4
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by BobtheBob View Post
    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= .
    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 $\displaystyle a^2+b^2+c^2=~?$
    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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  5. #5
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by Plato View Post
    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 $\displaystyle \sqrt{a^2+b^2+c^2}$

    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.
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  6. #6
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by BobtheBob View Post
    $\displaystyle \color{red}22=2(ab+ac+bc)$
    $\displaystyle \color{blue}24=4(a+b+c)$
    Look at what you already know.
    $\displaystyle \color{red}{2ab+2ac+2bc=22}~\&~\color{blue}{a+b+c= 6}$

    So what is $\displaystyle a^2+b^2+c^2~?$
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  7. #7
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    Re: Diagonal of a Cuboid

    Oh right so $\displaystyle a^2+b^2+c^2=36$?!
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by BobtheBob View Post
    Oh right so $\displaystyle a^2+b^2+c^2=36$?!
    Absolutely not.
    $\displaystyle (a+b+c)^2=36$
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  9. #9
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by Plato View Post
    Absolutely not.
    $\displaystyle (a+b+c)^2=36$
    Oh right of course!

    So

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

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


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

    $\displaystyle a^2+b^2+c^2+2ab+2ac+2bc=36$
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  10. #10
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by Plato View Post
    Absolutely not.
    $\displaystyle (a+b+c)^2=36$
    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?)
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