Problem 3. In the rectangle $A B C D$, with a perimeter of $60 \mathrm{~cm}$, it is given that $\overline{B C}=$ $\frac{2}{3} \overline{A B}$. On side $A B$, there is a point $E$ such that $\overline{A E}=\frac{1}{3} \overline{A B}$, and on side $B C$, there is a point $F$ such that $\overline{B F}=\frac{2}{3} \overline{B C}$. If point $G$ is the midpoint of segment $A D$, what is the area of triangle $E F G$?
Solution. Let $\overline{A B}=a$ and $\overline{B C}=b$. From the conditions of the problem, it follows that $2 a+2 b=60$ and $b=\frac{2}{3} a$, hence $2 a+2 \cdot \frac{2}{3} a=60$. Therefore,

$$
\frac{10}{3} a=60, \text { i.e. } a=18 \text{ cm}, b=12 \text{ cm}
$$

Now $\overline{A E}=6 \text{ cm}, \overline{E B}=12 \text{ cm}$,

![](https://cdn.mathpix.com/cropped/2024_06_05_e235c78af50361de0ec1g-07.jpg?height=341&width=452&top_left_y=1860&top_left_x=1014)

$$
\overline{B F}=8 \text{ cm}, \overline{F C}=4 \text{ cm} \text { and } \overline{A G}=\overline{G D}=6 \text{ cm}
$$

We have

$$
P_{E F G}=P_{A B C D}-\left(P_{A E G}+P_{B F E}+P_{C D G F}\right)
$$

Moreover,

$$
P_{A B C D}=18 \cdot 12=216 \text{ cm}^{2}, P_{A E G}=\frac{6 \cdot 6}{2}=18 \text{ cm}^{2}, P_{B F E}=\frac{12 \cdot 8}{2}=48 \text{ cm}^{2}
$$

and the area of quadrilateral $C D G F$ will be calculated by drawing a line through point $F$ parallel to side $C D$. We get

$$
P_{C D G F}=P_{C D H F}+P_{H G F}=18 \cdot 4+\frac{18 \cdot 2}{2}=90 \text{ cm}^{2}
$$

Finally, $P_{E F G}=216-(18+48+90)=216-156=60 \text{ cm}^{2}$.