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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.

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    $\begingroup$ One straightforward way: use the median length formula to calculate the sides, then use Heron's formula. That said, there is likely some clever way to take advantage of the fact that $5,12,13$ is a pythagorean triple. $\endgroup$ – dxiv Jan 9 '17 at 5:24
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    $\begingroup$ @dxiv Oh, didn't know about that formula. Thanks $\endgroup$ – suomynonA Jan 9 '17 at 5:40
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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 $$[\triangle{ABC}]=3\times[\triangle{BXC}]=\color{red}{90}$$

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  • $\begingroup$ How do you know $\angle XBD$ is $90^\circ$? $\endgroup$ – suomynonA Jan 9 '17 at 5:36
  • $\begingroup$ @suomynonA: $\angle{XBD}$ is not equal to $90^\circ$. I got $\angle{BXD}=90^\circ$ from the fact that $BD^2=BX^2+DX^2$. This is the converse of the Pythagorean theorem. $\endgroup$ – mathlove Jan 9 '17 at 5:41
  • $\begingroup$ @suomynonA -- The $90^\circ$ angle is $\angle BXD$, not $\angle XBD$. Check the side lengths of triangle $BXD$. $\endgroup$ – quasi Jan 9 '17 at 5:42
  • $\begingroup$ Oh sorry I read it wrong, also in the diagram I drew $\angle BXC$ looks about $45^\circ$, lol $\endgroup$ – suomynonA Jan 9 '17 at 5:42

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