# Math Help - pythag problem

1. ## 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

2. ## Re: pythag problem

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

3. ## Re: pythag problem

Thanks for the reply , but stilll stuck

4. ## Re: pythag problem

Please indicate the last claim of the proof that you understand and the first one that you don't.

5. ## Re: pythag problem

im struggling to see how edge L, has length a/sqrt 2

6. ## Re: pythag problem

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.

7. ## Re: pythag problem

i get the pythag , just not the ending

8. ## Re: pythag problem

why do we divide both sides by 2?

9. ## 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?

10. ## Re: pythag problem

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.

11. ## Re: pythag problem

thanks , all makes sense :0 )