# pythag problem

• Jul 21st 2012, 09:52 AM
davellew69
pythag problem
Hi all, I have an answer which is puzzeling me

if L is the length of the shorter edge, by pythag theorem, 2L^2 =a^2, so L^2=1/2a^2

Hence the shorter edge L, has length a/sqrt 2

if it helps the picture is a right angled triangle with a hyp of length given as a, and the two shorter lenghts given as L

what I cant get is why the shorter edge L, has length a/sqrt 2

Dave
• Jul 21st 2012, 10:28 AM
emakarov
Re: pythag problem
Quote:

Originally Posted by davellew69
what I cant get is why the shorter edge L, has length a/sqrt 2

Because 2L^2 =a^2.
• Jul 21st 2012, 10:30 AM
davellew69
Re: pythag problem
Thanks for the reply , but stilll stuck
• Jul 21st 2012, 10:31 AM
emakarov
Re: pythag problem
Please indicate the last claim of the proof that you understand and the first one that you don't.
• Jul 21st 2012, 10:33 AM
davellew69
Re: pythag problem
im struggling to see how edge L, has length a/sqrt 2
• Jul 21st 2012, 10:42 AM
emakarov
Re: pythag problem
Quote:

Originally Posted by davellew69
im struggling to see how edge L, has length a/sqrt 2

Sorry, you did not specify the last claim of the proof that you understand. This must mean that you don't understand why L^2 + L^2 = a^2. This is an instance of the Pythagoras theorem because we have a right-angled triangle with two sides of length L and the hypotenuse of length a.

So, we have 2L^2 = a^2. Divide both sides by 2. Now, take the square root of both sides.

I am just repeating the proof you gave. I am still curious to know which exactly part of it you don't understand.
• Jul 21st 2012, 11:00 AM
davellew69
Re: pythag problem
i get the pythag , just not the ending
• Jul 21st 2012, 11:01 AM
davellew69
Re: pythag problem
why do we divide both sides by 2?
• Jul 21st 2012, 11:13 AM
emakarov
Re: pythag problem
Look, is it too hard for you to say: "This equality I understand, but how this one follows from it I don't understand."
We have the following theorems about square root:

(1) If $x\ge0$, then $\sqrt{x^2}=x$.
(2) If $x\ge0$ and $y>0$, then $\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$.

So, we have the following derivation.

1. From the Pythagoras theorem, $2L^2=a^2$. Divide both sides by 2.
2. $L^2=\frac{a^2}{2}$. Take square root of both sides.
3. $\sqrt{L^2}=\sqrt{\frac{a^2}{2}}$. Apply (2)
4. $\sqrt{L^2}=\frac{\sqrt{a^2}}{\sqrt{2}}$. Apply (1)
5. $L=\frac{a}{\sqrt{2}}$.

Could you say which of the facts (1), (2) and steps 1-5 were not clear to you?
• Jul 21st 2012, 11:19 AM
emakarov
Re: pythag problem
Quote:

Originally Posted by davellew69
why do we divide both sides by 2?

Because in the final claim $L=\frac{a}{\sqrt{2}}$, L is alone in the left-hand side, but in the Pythagoras theorem $2L^2=a^2$, $L^2$ is multiplied by 2. We are trying to solve the latter equation for L, so we are trying to isolate L.
• Jul 21st 2012, 11:24 AM
davellew69
Re: pythag problem
thanks , all makes sense :0 )