Posted by Raymond on Sunday, September 11, 2011 at 3:29pm.
How would you solve for X if X was the hypotenuse of a 454590 triangle and the other two sides were Square Root of 11?

Math  MathMate, Sunday, September 11, 2011 at 3:44pm
Use Pythagoras theorem:
hypotenuse² = side1²+side2²l
=(√11)²+(√11)²
= 11 + 11
= 22
Therefore
hypotenuse = √22

Math  Raymond, Sunday, September 11, 2011 at 3:49pm
Thank you very much! I thought it was 2 Square Roots of 11. You saved me.

Math  MathMate, Sunday, September 11, 2011 at 7:14pm
The method I have shown above is more general, and applies to any righttriangle.
For a 454590 triangle, you were close with the answer. A shortcut would be to multiply each side by √2, which makes:
√2 * √11
=√22.
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