Triangle: Find area given 2 side lengths and that two medians are perpendicular Question: In triangle ABC, AC=6 and BC=8. It is also known that the median drawn from the vertices A and B are perpendicular to each other. Find the area of triangle ABC. 
I realize that because the medians are perpendicular, right triangles are formed within triangle ABC. However, I cannot figure out how to calculate the length of the medians using the fact that there are right triangles. 
 A: Let $A'$ be the midpoint of $BC$; let $B'$ be the midpoint of $AC$. Let $M$ be the intersection point of the medians. We will find the lengths of the medians $AA'$ and $BB'$ from the fact that the medians are divided by the intersection point $M$ forming the ratio $2/1$, that is,
$$
{AM\over A'M} = {BM\over B'M} = 2.
$$
Suppose that 
$${1\over3}AA'=MA'=x, \qquad {1\over3}BB'=MB'=y.$$
By the Pythagorean theorem, from the right triangles $AMB'$ and $BMA'$ we have the system of equations
$$
4x^2 + y^2 = 9,
$$
$$
x^2 + 4y^2 = 16.
$$
Solving these equations we find
$$
x = {2\over\sqrt3}, \qquad 
y = \sqrt{11\over3}.
$$
Therefore, the medians are 
$$
AA' = 3x = {2\sqrt3}, \qquad 
BB' = 3y = \sqrt{33}.
$$
We can find the third side, $AB$, from the right triangle $AMB$:
$$
AB^2 = AM^2 +BM^2 = 4x^2 + 4y^2 = 20.
$$
So we now know all tree sides of triangle $ABC$:
$$
a = BC = 8, \qquad b=AC=6, \qquad c = AB = \sqrt{20}.
$$
Finally, we can find the area of $ABC$ using an alternative form of Heron's formula:
$$
\mbox{Area}(ABC)=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
=4\sqrt{11}\approx13.266.
$$
