Calculate area of triangle? What is the area of triangle in which two of its medians are 9 cm and 12 cm long intersect at the right angles?
I tried this but could not get to the answer.
Does the triangle will also be right angle if the median intersect at right triangle?
Thanks in advance.
 A: 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.
A: Consider the triangle below: $|\overline{AC}|=2|\overline{AE}|$ and $|\overline{AB}|=2|\overline{AD}|$.
By SAS (side-angle-side), $\triangle ADE\sim\triangle ABC$. Therefore, $\overline{DE}||\overline{BC}$ and $|\overline{BC}|=2|\overline{DE}|$.
Since $\angle DEB=\angle EBC$ and $\angle EDC=\angle DCB$, ASA (angle-side-angle) yields $\triangle FED\sim\triangle FBC$.
Thus, $|\overline{FB}|=2|\overline{EF}|$ and $|\overline{FC}|=2|\overline{DF}|$. Therefore, the lengths are as listed. Since all the angles at $F$ are right, we get that $|\triangle BEC|=\frac128\cdot9=36$. The altitude of $\triangle ABC$ is twice that of $\triangle BEC$, so $|\triangle ABC|=72$.
$\hspace{3.5cm}$
Now that I've posted this, I see that it is the same as André Nicolas' second method (on the other internal triangle). I will leave it because of the diagram.
