Interesting question. Let the 2 unknown sides be x and x+1.
Then (11 m)^2 + x^2 = sum of the squares of the 2 shortest sides
= (x+1)^2
121 + x^2 = x^2 + 2x + 1. Then 121 = 2x + 1, and 2x = 120, or x = 60.
Then the hyp. has length 60+1= 61.
We must check these results. Using the Pyth. Thm. (a^2 + b^2 = c^2),
11^2 + 60^2 = 61^2
121 + 3600 = 3721 This is true, so our answers are correct.
The longer side is 60 and the hyp. has length 61.
