Let $X$ be the centroid of $\Delta ABC$, and $AX=5$, $BX=12$, $CX=13$. Find the area of $\Delta ABC$.
Some properties of the centroid that I know:
- The centroid divides the medians of a triangle in a $2:1$ ratio.
- The resulting six triangles formed after drawing the three medians of a triangle have equal area.
I don't know how to approach or solve this problem.
Take $D$ on the line $AX$ such that $X$ is the midpoint of the line segment $AD$.
Then, we see that the quadrilateral $BXCD$ is a parallelogram and that $BX=12,DX=AX=5,BD=XC=13$ with $BD^2=BX^2+DX^2$.
So, $[\triangle{BXC}]=[\triangle{BDC}]=[\triangle{BDX}]=\frac 12\times 12\times 5=30$ from which we have $$[\triangle{ABC}]=3\times[\triangle{BXC}]=\color{red}{90}$$