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Math Help - 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 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~?
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  3. #3
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by Plato View Post
    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= .

    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 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}
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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 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.

    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.
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  6. #6
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by BobtheBob View Post
    \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~?
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  7. #7
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    Re: Diagonal of a Cuboid

    Oh right so a^2+b^2+c^2=36?!
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  8. #8
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by BobtheBob View Post
    Oh right so a^2+b^2+c^2=36?!
    Absolutely not.
    (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.
    (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
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  10. #10
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    Re: Diagonal of a Cuboid

    Quote Originally Posted by Plato View Post
    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?)
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