On the sides $A B$ and $A C$ of triangle $A B C$, whose area is $36 \mathrm{~cm}^{2}$, points $M$ and $K$ are taken such that $A M / M B=1 / 3$, and $A K / K C=2 / 1$. Find the area of triangle $A M K$.

#
Draw $B K$.

## Solution

Connect points $B$ and $K$.

$$
\begin{aligned}
& S(A B K)=(2 / 3) \cdot S(A B C) \\
& S(A M K)=(1 / 4) \cdot S(A B K)
\end{aligned}
$$

Therefore,

$$
\begin{gathered}
S(A M K)=(2 / 3) \cdot(1 / 4) \cdot S(A B C)=(1 / 6) \cdot S(A B C)= \\
=(1 / 6) \cdot 36=6 .
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_34fd2032e75750f56a92g-32.jpg?height=375&width=826&top_left_y=2088&top_left_x=616)

## Answer

$6 \mathrm{~cm}^{2}$.

Submit a comment