Area of a triangle formed by mid-points of two sides and the other vertex

Question

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!

Answer 1

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.

Answer 2

Draw also the median from $C$. Observe that the area of all six triangles are equal. Find the total area using Heron's Formula and divide it by 6 to obtain the desired area.

Answer 3

Altitude from $M$ to $AB$ is $\frac{1}{2}$ of altitude from $C$ to $AB$.

So $\triangle AMB = \frac{1}{2} \triangle ABC$.

Now taking base as $AM$, altitude from $O$ to $AM$ is $\frac{1}{3}$ of altitude from $B$ to $AM$.
(as $OM$ is $\frac{1}{3}$ of $BM$).

So $\triangle AOM = \frac{1}{3} \triangle AMB = \frac{1}{6} \triangle ABC$.

You can find area of $\triangle ABC$ using its side lengths.