2. (17 points) The medians drawn from vertices $A$ and $B$ of triangle $ABC$ are perpendicular to each other. Find the area of the square with side $AB$, if $BC=36, AC=48$.

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# Answer: 720

Solution. Let $D$ be the midpoint of $B C$, $E$ be the midpoint of $A C$, and $M$ be the point of intersection of the medians. Let $M D=a$, $M E=b$. Then $A M=2 a$, $B M=2 b$. From the right triangles $B M D$ and $A M E$, we have $a^{2}+4 b^{2}=B D^{2}=18^{2}$ and $4 a^{2}+b^{2}=A E^{2}=24^{2}$. Adding these equations and dividing by 5, we get $a^{2}+b^{2}=180$. Therefore, $A B^{2}=A M^{2}+B M^{2}=4\left(a^{2}+b^{2}\right)=720$.