Let our triangle be $ABC$, and let the two medians be $AM$ (of length $12$) and $BN$ (of length $9$), meeting at the centroid $X$. Recall that the centroid of a triangle divides each median in the proportions $2:1$. It follows that $\triangle ABX$ is right-angled at $X$ and has "legs" $8$ and $6$. Thus $\triangle ABX$ has area $24$.
But the area of $\triangle ABX$ is one-third of the area of $\triangle ABC$, since the two triangles share a base $AB$, and the median from $C$ is split by $X$ in the proportion $2:1$, which implies that that the height of $\triangle ABC$ is $3$ times the height of $\triangle ABX$. So $\triangle ABC$ has area $72$.
Another way: Here is a simpler but less symmetric way. Let $AM=12$ and $BN=9$. Then $BX=6$, so $\triangle ABM$ has area $(12)(6)/2=36$. It follows that $\triangle ABC$ has area $72$.
Another way: Draw all the medians. They divide the triangle into $6$ triangles of equal area. But one of them, $\triangle MXB$, is right-angled at $X$ and has legs $4$ and $6$, so area $12$.
Remark: Part of your question asked whether the original triangle is right-angled. It isn't. Calculation shows that $\triangle ABC$ with $AB=10$, $BC=4\sqrt{13}$, and $CA=2\sqrt{73}$ is the only triangle that has medians of length $12$ and $9$ meeting at right angles. This triangle is not right-angled.