COMMON LETS FIND THE LONGEST ONE AGAIN

Let $a$ and $b$ be the legs and $c$ be the hypotenuse of the right triangle.
Clearly $\dfrac{ab}2 = 1386$ , $a+b+c = 198$ and $a^2 + b^2 = c^2$ . (The last one is the Pythagorean theorem .)

Now, $2ab = 4\cdot 1386$ and $a^2 + b^2 + 2ab = \left(a+b\right)^2 = \left(198-c\right)^2 = c^2 - 2 \cdot 198c + 198^2$ ,
which imply that $c^2 + 4 \cdot 1386 = c^2 - 2 \cdot 198c + 198^2 \implies c = \dfrac{198^2 - 4 \cdots 1386}{2\cdot198} = 99 - 14 = \color{#3D99F6}{85}$ .

Hence, the answer is $\boxed{\color{#3D99F6}{85}}$ .