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 Post subject: Manipulation of square roots in triangle problems
PostPosted: Wed Jul 08, 2009 10:35 am 
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Joined: Wed Jul 08, 2009 10:22 am
Posts: 1
Hi,

I have a question around the manipulation of the square roots for isosceles triangles.
In example 1 on the triangles page, the hypotenuse is root2 x length of either of the other sides. As the hyp is geven as 20, i make the answer 20/root 2, yet i know this is wrong as the answer given is very different (and difficult to display here). Could you give me a more detailed run through of the explanation, particularly on how you get from L = (1/root 2) x 20 to the next stage.

Thanks


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 Post subject: Re: Manipulation of square roots in triangle problems
PostPosted: Sun Jul 26, 2009 11:19 am 
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Joined: Mon Apr 06, 2009 5:44 pm
Posts: 81
If the hypotenuse is equal to 20 and it is also equal to sqrt(2) times the length of each side then, we need to set up the following equation

20 = sqrt(2) x where x represents the length of either of the two sides. Now, we just solve for x:

x = 20 / sqrt (2) = 10sqrt(2)

I hope this helps,

Steve


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