B2. The difference in the perimeters of two squares is $8 \mathrm{~cm}$, and the difference in their areas is $16 \mathrm{~cm}^{2}$. Calculate the sum of their areas.
B2. The difference in perimeters should be $4 a-4 b=8 \mathrm{~cm}$, and the difference in areas $a^{2}-b^{2}=16 \mathrm{~cm}^{2}$. From the first equation, we get $a-b=2$, and from the second, due to $(a-b)(a+b)=16$, we also get $a+b=8$. From this, we arrive at $a=5 \mathrm{~cm}$ and $b=3 \mathrm{~cm}$. The sum of both areas is $5^{2}+3^{2}=34 \mathrm{~cm}^{2}$.

Writing the difference in perimeters $4 a-4 b=8$. 1 point

Writing the difference in areas $a^{2}-b^{2}=16$ 1 point

Correct solving of the system 1 point

Solutions $a=5 \mathrm{~cm}$ and $b=3 \mathrm{~cm}$. $1+1$ points

Calculated sum of areas: $S_{1}+S_{2}=34 \mathrm{~cm}^{2}$ 1 point