Problem 2. The sides of a right triangle are consecutive natural numbers. The perimeter of the right triangle is $12 \mathrm{dm}$. A square is constructed on each side of the triangle. Determine the area of the resulting figure.
Solution. The sides of the triangle are consecutive natural numbers, i.e., they are $x-1, x$, and $x+1$. Since the perimeter of the triangle is $12 dm$, we get $x-1+x+x+1=12 dm$, from which solving the equation gives $3x=12 dm$, i.e., $x=4 dm$. It follows that the sides of the triangle have lengths $3 \mathrm{dm}, 4 \mathrm{dm}$, and $5 \mathrm{dm}$. The area of the right-angled triangle is $P_{\Delta}=\frac{3 \cdot 4}{2} \mathrm{dm}^{2}=6 \mathrm{dm}^{2}$, and the areas of the squares are:

$$
P_{1}=3 \cdot 3 \mathrm{dm}^{2}=9 \mathrm{dm}^{2}, P_{2}=4 \cdot 4 \mathrm{dm}^{2}=16 \mathrm{dm}^{2} \text { and } P_{3}=5 \cdot 5 \mathrm{dm}^{2}=25 \mathrm{dm}^{2}
$$

The area of the resulting figure is

$$
P=P_{1}+P_{2}+P_{3}+P_{\Delta}=9+16+25+6=56 \mathrm{dm}^{2}
$$