TITLE: Calculate area of triangle?
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.