Let the medians $AK$ and $BM$ of a triangle $ABC$ intersect at $O$, $AB=13,BC=14,CA=15$. Find the area of $\triangle AOM$.
The lengths of the medians are given by:$$m_a=AK=\dfrac12\sqrt{2b^2+2c^2-a^2},\\m_b=BM=\dfrac12\sqrt{2a^2+2c^2-b^2}.$$ So $AM=\dfrac{15}{2}$ and we can use the fact that the centroid $O$ divides each median into two parts, which are always in the ratio $2:1$ to find $AO$ and $MO$. Then we can find the area by Heron's formula. I am not sure this is the easiest approach. What do you think? Thank you in advance!
Area of a triangle formed by mid-points of two sides and the other vertex
centroideuclidean-geometrygeometrytriangles
Best Answer
Since $O$ is the centroid, the area of $\triangle AOM$ is $\frac16$ of that of $\triangle ABC$.
The area of $\triangle ABC$ is easily found by Heron's Formula; hence the area of $\triangle AOM$ is easily found as well.