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