3. The difference in the perimeters of two squares is $8 \mathrm{~cm}$, and the difference in their areas is $16 \mathrm{~cm}^{2}$. What is the sum of their areas? Write down the answer.
3. From the difference in the perimeters of the squares, we determine $a_{1}-a_{2}=2$. From the difference in the areas of the squares, we get $a_{1}^{2}-a_{2}^{2}=16$, which can be factored as $\left(a_{1}+a_{2}\right)\left(a_{1}-a_{2}\right)=16$. Since $a_{1}-a_{2}=2$, it follows that $a_{1}+a_{2}=8$. From the system of equations $a_{1}-a_{2}=2, a_{1}+a_{2}=8$, we obtain the solutions $a_{1}=3 \mathrm{~cm}$ and $a_{2}=5$ $\mathrm{cm}$. We calculate the areas of both squares $a_{1}^{2}=9 \mathrm{~cm}^{2}$ and $a_{2}^{2}=25 \mathrm{~cm}^{2}$. The sum of the areas is 34 $\mathrm{cm}^{2}$.

Determination $4 a_{1}-4 a_{2}=8 \Rightarrow a_{1}-a_{2}=2$ 1 point

Considering the difference of the squares of the areas

![](https://cdn.mathpix.com/cropped/2024_06_07_59669f975316a35884bfg-11.jpg?height=60&width=1628&top_left_y=1883&top_left_x=217)

![](https://cdn.mathpix.com/cropped/2024_06_07_59669f975316a35884bfg-11.jpg?height=49&width=1628&top_left_y=1940&top_left_x=217)

Calculation:

![](https://cdn.mathpix.com/cropped/2024_06_07_59669f975316a35884bfg-11.jpg?height=57&width=1628&top_left_y=2033&top_left_x=217)

![](https://cdn.mathpix.com/cropped/2024_06_07_59669f975316a35884bfg-11.jpg?height=52&width=1636&top_left_y=2087&top_left_x=210)

![](https://cdn.mathpix.com/cropped/2024_06_07_59669f975316a35884bfg-11.jpg?height=60&width=1639&top_left_y=2129&top_left_x=206)