$3+$ [The area of a figure is equal to the sum of the areas of the figures into which it is divided]

Points $E, F, M$ are located on the sides $A B, B C, A C$ of triangle $A B C$, respectively. Segment $A E$ is one third of side $A B$, segment $B F$ is one sixth of side $B C$, and segment $A M$ is two fifths of side $A C$. Find the ratio of the area of triangle $E F M$ to the area of triangle $A B C$.
$S_{\triangle A E M}=\frac{A E}{A B} \cdot \frac{A M}{A C}$

## Solution

Let $S$ be the area of triangle $A B C$. Then

$$
S_{\triangle \mathrm{BEF}}=\frac{B E}{A B} \cdot \frac{B F}{B C} S_{\triangle \mathrm{ABC}}=\frac{2}{3} \cdot \frac{1}{6} S=\frac{1}{9} S
$$

Similarly, $S_{\Delta \mathrm{AEM}}=\frac{2}{15} S$ and $S_{\Delta \mathrm{CMF}}=\frac{1}{2} S$. Therefore,

$$
S_{\triangle \mathrm{EFM}}=S-\frac{1}{9} S-\frac{2}{15} S-\frac{1}{2} S=\frac{23}{90} S
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-55.jpg?height=418&width=914&top_left_y=1429&top_left_x=572)

## Answer