On the side $A D$ of the rhombus $A B C D$, a point $M$ is taken such that $M D=0.3 A D$ and $B M=M C=11$. Find the area of triangle $B C M$.

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Let $A D=10 x, M D=3 x$. If $K-$ is the projection of point $M$ on side $B C$, then $B K=K C=5 x, M K^{2}=B M^{2}-B K^{2}=$ $121-25 x^{2}$.

If $F$ is the projection of point $B$ on side $A D$, then $A F=A M-F M=A M-B K=2 x, B F^{2}=A B^{2}-A F^{2}=100 x^{2}-4 x^{2}=$ $96 x^{2}$, and since $B F=M K$, then $121-25 x^{2}=96 x^{2}$

From this, $x=1, B C=10 x=10, K M^{2}=121-25=96$.

Therefore, $S_{B C M}=1 / 2 B C \cdot K M=20 \sqrt{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-03.jpg?height=385&width=603&top_left_y=-1&top_left_x=733)

## Answer

$20 \sqrt{6}$.