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.
We'll derive the equation using the fact:
$$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, \quad (I)$$
Using the angle bisector theorem we get:
$$BP=\frac{ac}{a+b},\quad (1)$$
$$BR=\frac{ac}{b+c}, \quad (2)$$
$$CR=\frac{ab}{b+c},\quad (3)$$
$$CQ=\frac{ab}{a+c},\quad (4)$$
$$AQ=\frac{bc}{a+c},\quad (5)$$
and
$$AP=\frac{bc}{a+b}. \quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=\frac{1}{2}ab\sin\gamma, \quad (7)$$
$$A_{PBR}=\frac{1}{2}BP\cdot BR\sin\beta, \quad (8)$$
$$A_{RCQ}=\frac{1}{2}CR\cdot CQ\sin\gamma, \quad (9)$$
and
$$A_{QAP}=\frac{1}{2}AQ\cdot AP\sin\alpha. \quad (10)$$
Let $R$ be the circumradius, we know that:
$$\sin \alpha = \frac{a}{2R}, \quad (11)$$
$$\sin \beta = \frac{b}{2R}, \quad (12)$$
$$\sin \gamma = \frac{c}{2R}, \quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get:
$$A_{PQR}=\frac{1}{2} \cdot \frac{abc}{2R}-\frac{1}{2} \frac{a^2c^2b}{(a+b)(b+c)2R}-\frac{1}{2} \cdot \frac{a^2b^2c}{(b+c)(a+c)2R}-\frac{1}{2} \cdot \frac{b^2c^2a}{(a+b)(a+c)2R}, \Rightarrow$$
$$A_{PQR}=\frac{abc}{4R}[1-\frac{ac}{(a+b)(b+c)}-\frac{ab}{(b+c)(a+c)}-\frac{bc}{(a+b)(a+c)}], \Rightarrow$$
$$A_{PQR}=\frac{abc}{2R}[\frac{abc}{(a+b)(b+c)(a+c)}], \Rightarrow$$
$$A_{PQR}=A_{ABC}[\frac{2abc}{(a+b)(b+c)(a+c)}]$$
Using Heron's formula we are done.
Best Answer
Since $LO$ is both median and the bisector of $\angle NLM$, $\triangle LMN$ is isosceles and $ LO \perp MN$.
So $MO = NO= \sqrt{5^2-3^2\ }=4$ and area $\triangle LMN = 4\times 3 =12 \text{cm}^2$
To see that $\triangle LMN$ is isosceles, consider that $\angle LON = 180°-\angle LOM$. Flip $\triangle LON$ to superimpose $\angle NLO$ and $\angle MLO$. Then the only way to make $|MO|=|NO|$ is to have $\angle LON = \angle LOM =\perp$