Problem 3. A rectangle is divided into six squares, as shown in the diagram. Determine the area of the rectangle if the side of the smallest square is $2 \mathrm{~m}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_4fa80d9916883368a7a6g-6.jpg?height=199&width=216&top_left_y=1799&top_left_x=1209)
Solution. Let $x$ be the side of square $ABCD$. Then the side of square DHFE is $x+2$, i.e., $\overline{DE}=\overline{EF}=x+2$. Note that $\overline{BS}=\overline{NM}=x-2$. Thus, we have $\overline{BN}=2x-4$. Therefore,

$$
\overline{AN}=3x-4 \text{ and } \overline{AE}=2x+2
$$

For $\overline{FG}$, we have

$\overline{FG}=\overline{AN}-\overline{EF}=(3x-4)-(x+2)=2x-6$

Similarly, we get

$\overline{MG}=\overline{AE}-\overline{NM}=(2x+2)-(x-2)=x+4$

From $\overline{FG}=\overline{MG}$, we obtain the equation $2x-6=x+4$, from which we get

![](https://cdn.mathpix.com/cropped/2024_06_05_4fa80d9916883368a7a6g-7.jpg?height=487&width=546&top_left_y=266&top_left_x=938)
$x=10$. Thus, we have

$$
\overline{AN}=3 \cdot 10-4=26 \text{ and } \overline{AE}=2 \cdot 10+2=22 \text{. }
$$

Therefore, the area of the rectangle is

$$
P=22 \cdot 26=572 m^{2}
$$