2. Find the area of the triangle if it is known that its medians $C M$ and $B N$ are 6 and 4.5 respectively, and $\angle B K M=45^{\circ}$, where $K$ is the point of intersection of the medians.
Solution. The given triangle $ABC$ is shown in the diagram, where the medians $CM$ and $BN$ are 6 and 4.5, respectively, $\angle BKM=45^{\circ}$, and $K$ is the point of intersection of the medians. It is easy to prove that the medians of a triangle divide it into 6 equal-area triangles. Therefore, to find the area of $\triangle ABC$, it is sufficient to find the area of $\triangle BKM$ and multiply it by 6.

By the property of the point of intersection of the medians, $\frac{BK}{BN}=\frac{2}{3}$ and $\frac{MK}{CM}=\frac{1}{3}$, so $BK=\frac{2}{3}BN=\frac{2}{3} \cdot 4.5=3$, $MK=\frac{1}{3}CM=\frac{1}{3} \cdot 6=2$. Then $S_{\triangle BKM}=\frac{1}{2} BK \cdot MK \cdot \sin \angle BKM=\frac{1}{2} \cdot 3 \cdot 2 \cdot \sin 45^{\circ}=\frac{3 \sqrt{2}}{2}$. Therefore,

$S_{\triangle ABC}=6 \cdot \frac{3 \sqrt{2}}{2}=9 \sqrt{2}$.

Answer: $9 \sqrt{2}$.

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