Solving area of a triangle where medians are perpendicular.

Question

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.

Answer 1

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 $200$.

Answer 2

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