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Medians $\overline{AD}$ and $\overline{BE}$ of a $\triangle ABC$ are perpendicular. If $AD= 15$ and $BE = 20$, then what is the area of $\triangle ABC$?

Note: A lot of my work can have inaccuracies and is based off a diagram. It is very helpful if you draw a diagram first

Let's call the centroid of $\triangle ABC$, $G$. Since we know that $\frac{AG}{GD}=\frac{2}{3}$, we have that $AG=10$. We also have that $GD=\frac{20}{3}$. This gives us $\frac{10\sqrt{13}}{3}$ as the length of $AE$. We can draw a perpendicular from $D$ to a point on $AC$ (we'll call this point $H$) such that $\angle{ADH}=90^\circ$. We can use similar triangles to determine that the side length of $AH=5\sqrt{13}$. From there, we can once again use similar triangles to find that $HC=\frac{20}{27}$. From here, I don't know what to do.

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Let $AD$ and $BD$ meet at $G$ = gravity center. Remember that $AG:GD =2:1$ so $AG=10$.

Area of the triangle $ABD$ which is half of the area of whole triangle $ABC$ is $${EB \cdot AG \over 2} = {20 \cdot 10\over 2} =100$$ so the whole triangle has area $200$.

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  • $\begingroup$ How did u directly obtain this, did you use some theorem $\endgroup$ – King Tut Mar 4 '18 at 19:13
  • $\begingroup$ No I was not having that problem, I was asking how you get $\text{Ar}(ABD)$ as what you got... $\endgroup$ – King Tut Mar 4 '18 at 19:50
  • $\begingroup$ Yes thats my trouble.. in $ABD$ height is $BG$ and base is $AD$. You used $EB\cdot AG$ but it works fine here because the centroid ratio gets adjusted. Is it some general rule? $\endgroup$ – King Tut Mar 4 '18 at 19:55
  • $\begingroup$ If you have $\angle AGB = \phi$ then $$ Area(ABC)= 3\cdot {AG\cdot BG \cdot \sin (\phi) \over 2}$$ $\endgroup$ – Aqua Mar 4 '18 at 19:57
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These following equalities are using the fact that area of triangle with same vertex and equal base are same.

$$\text{Ar}(ABC) = 2\text{Ar}(ABD) = 2[\tfrac{3}{2}\text{Ar}(AGB)] = 3\text{Ar}(AGB)$$

Now $\triangle AGB$ is right triangle with $AG = \frac{2AD}{3} = 10$ and $GB = \frac{2BG}{3} = \frac{40}{3}$.

Required area is $$3\text{Ar}(AGB)=\frac{3\cdot AG\cdot GB}{2} = 200$$

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