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.
Best Answer
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.