Problem#16

The hypotenuse is of length $10$ ; Pythagoras gives us: $a^2 + b^2 = 10^2 = 100$

The area of the triangle is $10$ , which gives: $\frac{1}{2} a b = 10$ or $ab=20$

If we add $2ab$ to the first equation, we should get something recognisable: $a^2 + 2ab + b^2 = 100 + 2ab = 140\\ a^2 + 2ab + b^2 = (a+b)^2 = 140\\ a+b = \sqrt{140}$

This is the only possible value for $a+b$ , from which it is the maximum. The maximum perimeter is, therefore, $a+b+10 = 10+\sqrt{140}$