



 
Hi Audrey, I started by saying suppose the hypotenuse has length $2x  1$ units. Then by Pythagoras Theorem \[(x  2)^2 + (x + 5)^2 = (2x  1)^2\] I then expanded this equation, simplified it and solved the resulting quadratic equation for $x.$ I found two solutions, one with $x$ positive and the other with $x$ negative. I then checked the positive value and found that the lengths did form a Pythagorean Triple which gave me a possible length for the hypotenuse. The negative value for $x$ gives $x  2$ as negative and hence it can't be the length of the side of a triangle. Now suppose the hypotenuse is of length $x + 5$ and perform the analogous procedure. Finally let the hypotenuse have length $x  2$ and repeat the process. Penny  


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