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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.

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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: $$ \mbox{Area}(ABC)=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} =4\sqrt{11}\approx13.266. $$

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