We present five different solutions and outline a sixth and seventh one. In what follows, let $a=BC$, $b=CA$, $c=AB$ as usual, and denote by $r$ and $R$ the inradius and circumradius. Let $s=\frac{1}{2}(a+b+c)$. In the first five solutions we will only prove that $\angle AIO=90^{\circ} \Longrightarrow b+c=2a$. Let us see how this solves the problem. This lemma implies that $s=216$. If we let $E$ be the foot of $I$ on $AB$, then $AE=s-BC=72$, consequently the inradius is $r=\sqrt{97^{2}-72^{2}}=65$. Finally, the area is $sr=216 \cdot 65=14040$.