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Math Help - Prove the height given a diagonal is in d metres ... help please :(-

  1. #1
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    Prove the height given a diagonal is in d metres ... help please :(-

    Hi all,

    Am trying to help my autistic son with his review and one of the question is as per the attachment. I uploaded a .gif version of the question because am not sure whether I can enter the question here properly.

    The question goes as below:

    ===============

    The length of the diagonal of the base of the container is d metres.

    Show that the height is given by
    Prove the height given a diagonal is in d metres ... help please :(--math-help-01a.gif
    Prove the height given a diagonal is in d metres ... help please :(--math-help-01b.gif


    ===============

    I do not know whether am trying to solve the problem correctly or not. I tried to solve the problem the only way I know how which is using Pythagoras Theorem and doing

    d^2 = (2h+1)^2 + (h+3)^2

    And am stucked at

    d^2 = 5h^2 + 10h + 10

    Can anyone please advise whether am trying to solve this problem correctly or not. Any advise much appreciated. For the meantime, I'll try Google and see if I can find a similar question.

    Thanks in advance.
    Attached Thumbnails Attached Thumbnails Prove the height given a diagonal is in d metres ... help please :(--math-help-01.gif  
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  2. #2
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    Re: Prove the height given a diagonal is in d metres ... help please :(-

    Quote Originally Posted by bartholomew View Post
    Hi all,

    Am trying to help my autistic son with his review and one of the question is as per the attachment. I uploaded a .gif version of the question because am not sure whether I can enter the question here properly.

    The question goes as below:

    ===============

    The length of the diagonal of the base of the container is d metres.

    Show that the height is given by
    Click image for larger version. 

Name:	Math-Help-01a.gif 
Views:	7 
Size:	1.9 KB 
ID:	22757
    Click image for larger version. 

Name:	Math-Help-01b.gif 
Views:	7 
Size:	3.9 KB 
ID:	22756


    ===============

    I do not know whether am trying to solve the problem correctly or not. I tried to solve the problem the only way I know how which is using Pythagoras Theorem and doing

    d^2 = (2h+1)^2 + (h+3)^2

    And am stucked at

    d^2 = 5h^2 + 10h + 10

    Can anyone please advise whether am trying to solve this problem correctly or not. Any advise much appreciated. For the meantime, I'll try Google and see if I can find a similar question.

    Thanks in advance.
    Pythagoras is a good choice. Note that you have a quadratic in h so rearrange into the form ah^2+bh+c = 0 which is 5h^2 + 10h + (10-d^2) = 0

    Use the quadratic formula to solve for h. Hint: 5 = \dfrac{25}{5}
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  3. #3
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    Re: Prove the height given a diagonal is in d metres ... help please :(-

    Quote Originally Posted by e^(i*pi) View Post
    Pythagoras is a good choice. Note that you have a quadratic in h so rearrange into the form ah^2+bh+c = 0 which is 5h^2 + 10h + (10-d^2) = 0

    Use the quadratic formula to solve for h. Hint: 5 = \dfrac{25}{5}
    Hi,

    Thanks a lot for your reply. Very much appreciated. Am trying to answer a lot of the examples as much as I can so my son can review it when he wakes up tomorrow.

    Anyway, am not sure how to do a PM. Is that Private Message? I tried and it says I can't do a PM until my post reaches 10 -. Can I actually send a PM from the FORUM or do I have to cut-and-paste my post into the PM?

    Can't quite get the hint that you've given. But I tried using quadratic equation. So I have 5h^2 + 10h + 10 -d^2 = 0

    And using the quadratic equation formula, I end up with

     h = -1 +/- sqrt(20d^2-100)

    Thanks again for your response. Much appreciated.

    BTW, how do you "draw" +/- and the sqrt symbol in the symbol?
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  4. #4
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    e^(i*pi)'s Avatar
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    Re: Prove the height given a diagonal is in d metres ... help please :(-

    Quote Originally Posted by bartholomew View Post
    Hi,

    Thanks a lot for your reply. Very much appreciated. Am trying to answer a lot of the examples as much as I can so my son can review it when he wakes up tomorrow.

    Anyway, am not sure how to do a PM. Is that Private Message? I tried and it says I can't do a PM until my post reaches 10 -. Can I actually send a PM from the FORUM or do I have to cut-and-paste my post into the PM?
    Nah, post in the thread if it's something to do with the question. PM (which is private message) is if you wanted to chat about something more general or get in touch with me about something other than a question. Questions go in threads like this one.

    BTW, how do you "draw" +/- and the sqrt symbol in the symbol?
    [tex]\pm[/tex] is \pm and [tex]\sqrt{ax+b}[/tex] is \sqrt{ax+b}. You'll pick up latex tricks as you post

    Can't quite get the hint that you've given. But I tried using quadratic equation. So I have 5h^2 + 10h + 10 -d^2 = 0

    And using the quadratic equation formula, I end up with

     h = -1 +/- sqrt(20d^2-100)

    Thanks again for your response. Much appreciated.
    I'm not sure what happened to your 10 on the denominator, that should still be there on the second term. In the numerator you can factor out 20 (which I shall write as 4*5)

    h = -1 \pm \dfrac{\sqrt{4 \times 5(d^2-5)}}{10} = -1 \pm \dfrac{2 \sqrt{5(d^2-5)}}{10}

    Here is where the hint I suggested comes in. Write that 5 under the radical as \dfrac{25}{5} which is h = -1 \pm \dfrac{2 \sqrt{\dfrac{25}{5}(d^2-5)}}{10}

    As 25 =5^2 we can pull that 25 out of the radical: h = -1 \pm \dfrac{2 \times 5 \sqrt{\dfrac{d^2-5}{5}}}{10} = -1 \pm \sqrt{\dfrac{d^2-5}{5}

    QED
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    Re: Prove the height given a diagonal is in d metres ... help please :(-

    Quote Originally Posted by e^(i*pi) View Post
    Nah, post in the thread if it's something to do with the question. PM (which is private message) is if you wanted to chat about something more general or get in touch with me about something other than a question. Questions go in threads like this one.



    [tex]\pm[/tex] is \pm and [tex]\sqrt{ax+b}[/tex] is \sqrt{ax+b}. You'll pick up latex tricks as you post



    I'm not sure what happened to your 10 on the denominator, that should still be there on the second term. In the numerator you can factor out 20 (which I shall write as 4*5)

    h = -1 \pm \dfrac{\sqrt{4 \times 5(d^2-5)}}{10} = -1 \pm \dfrac{2 \sqrt{5(d^2-5)}}{10}

    Here is where the hint I suggested comes in. Write that 5 under the radical as \dfrac{25}{5} which is h = -1 \pm \dfrac{2 \sqrt{\dfrac{25}{5}(d^2-5)}}{10}

    As 25 =5^2 we can pull that 25 out of the radical: h = -1 \pm \dfrac{2 \times 5 \sqrt{\dfrac{d^2-5}{5}}}{10} = -1 \pm \sqrt{\dfrac{d^2-5}{5}

    QED


    Thanks for the updated response. I lost the 10 in the denominator 'coz I assumed it is supposed to disappear after I simplify the first term ... ouch ... I got a long way to go to learn these stuff ...

    That's a very good explanation. Thanks again for your help ... The Latex syntax are getting addictive and fun
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