3 Attachment(s)

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

Attachment 22757

Attachment 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.

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

Quote:

Originally Posted by

**bartholomew** 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

Attachment 22757 Attachment 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 $\displaystyle h$ so rearrange into the form $\displaystyle ah^2+bh+c = 0$ which is $\displaystyle 5h^2 + 10h + (10-d^2) = 0$

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

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

Quote:

Originally Posted by

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

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

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 $\displaystyle 5h^2 + 10h + 10 -d^2 = 0 $

And using the quadratic equation formula, I end up with

$\displaystyle 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?

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

Quote:

Originally Posted by

**bartholomew** 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.

Quote:

BTW, how do you "draw" +/- and the sqrt symbol in the symbol?

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

Quote:

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

And using the quadratic equation formula, I end up with

$\displaystyle 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)

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

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

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

QED

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

Quote:

Originally Posted by

**e^(i*pi)** 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 $\displaystyle \pm$ and [tex]\sqrt{ax+b}[/tex] is $\displaystyle \sqrt{ax+b}$. 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)

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

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

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

QED

(Bow)

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 (Headbang) ... 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