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.


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 $200$.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.