[Ratio of areas of triangles with a common angle]

On the sides $AB$ and $AC$ of triangle $ABC$, whose area is 50, points $M$ and $K$ are taken such that $AM: MB=1: 5$, and $AK: KC=3: 2$. Find the area of triangle $AMK$.

#
$$
S_{\triangle A M K}=\frac{A M}{A B} \cdot \frac{A K}{A C} S_{\triangle A B C}=\frac{1}{6} \cdot \frac{3}{5} \cdot 50=5
$$

## Answer

5.