Maximising Perimeter

The hypotenuse cannot be of length 12, as then the perimeter would obviously be too small.

Therefore, one of the other side has length 12. Say the hypotenuse has length a, and the other side has length b.

Then $a^{2}-b^{2}=144$

$(a-b)(a+b)=144$ .

For the perimeter to be maximised, a+b has to be maximised.

If $a+b=144$ , then $a-b=1$ , but then a and b are not integers.

If $a+b=72$ , then $a-b=2$ , so $a=37$ and $b=35$

Therefore, the maximum perimeter is 72+12=84.

Note: this problem could be generalised- What is the greatest perimeter of a right triangle having integer side lengths if one of the sides is of length x, where $x>2$ ?

If x is odd then the maximum is $x^{2}+x$

If x is even then the maximum is $\frac{x^{2}}{2}+x$