Area of triangle given two medians and one side.

In triangle $ABC$,point $D$ is the midpoint of $BC$and point $E$ is the midpoint of $AC$.If $AD=6$,$BE=9$,and $AC=10$ then find the area of triangle $ABC$.

I've solved this problem in two ways:

$1)$ by constructing the third median and calculating the area of one of the triangle determined by the intersection of the medians then multipy by six.

$2)$ by calculating the lengths of the other two unknown sides and then applying Heron's Formula.

Now what i am asking is if someone can come up with another solution based on geometric methods.

P.S:you may ask the why of this question and the answer would be that i like to see many creative solutions to one problem,and see what is the most beautiful one.

Let $BE$ amd $AD$ meet at G, the centroid. It is a well-known property of the centroid that it divides a median in the ratio $2:1$. Therefore $AG=4$ and $GE=3$. Therefore $AEG$ is a $3-4-5$ i.e. right-angled triangle whose area is $6$. Therefore the total area is $36$
• a typo maybe ,$GE=3$ – chenbai Sep 12 '15 at 4:52