Since the smaller squares have an area of $1 m^{2}$, each of them has a side length of $1 \mathrm{~m}$. From the figure, we conclude that $B C=2 m$ and $B H=3 m$.

![](https://cdn.mathpix.com/cropped/2024_05_01_ed653f4a7a838e105fd2g-08.jpg?height=459&width=725&top_left_y=1992&top_left_x=194)

Since $A B C D$ is a square, it follows that $B C=C D=A D=A B=2 m$. Since $C D E F$ is also a square, we have $C D=D E=2 m$. Again from the figure, we have: $A H=A B+B H=2+3=5, \quad J E=J A+A D+D E$ and $J A=A H$. It follows that $J E=5+2+2=9$. Since $E G=A H=5$, the dimensions of the land are $9 m$ in length by $5 m$ in width. Therefore, its area is $9 m \times 5 m=45 m^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_ed653f4a7a838e105fd2g-09.jpg?height=422&width=745&top_left_y=283&top_left_x=130)
2 - By tiling the larger rectangle with squares of $1 \mathrm{~m}^{2}$ in area, we obtain a rectangle ($B F G H$) formed by 12 squares of $1 \mathrm{~m}^{2}$ in area, two squares ($A B C D$ and $D C F E$) each formed by 4 squares of $1 \mathrm{~m}^{2}$ in area, and one square ($A H I J$) formed by 25 squares of $1 \mathrm{~m}^{2}$ in area. Therefore, the area requested is $12+4+4+25=45 \mathrm{~m}^{2}$.