6. It is known that the perimeter of one rectangle is greater than the perimeter of another rectangle. Compare the areas of these rectangles.



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6. The solution to this problem can be obtained by students as a result of conducting a computational experiment with various rectangles, for example:
a) $a=3 \text{~cm}, b=4 \text{~cm}$;

$$
\begin{aligned}
& a=4 \text{~cm}, b=2 \text{~cm} ; \\
& p_{2}=12 \text{~cm} ; \\
& S_{2}=8 \text{~cm}^{2}
\end{aligned}
$$

$$
\begin{aligned}
& p_{1}=14 \text{~cm} ; \\
& S_{1}=12 \text{~cm}^{2} ;
\end{aligned}
$$

$$
S_{1}>S_{2}
$$

b) $a=6 \text{~cm}, b=1 \text{~cm}$;

$$
\begin{aligned}
& a=4 \text{~cm}, b=2 \text{~cm} ; \\
& p_{2}=12 \text{~cm} \\
& S_{2}=8 \text{~cm}^{2}
\end{aligned}
$$

$$
\begin{aligned}
& p_{1}=14 \text{~cm} \\
& S_{1}=6 \text{~cm}^{2}
\end{aligned}
$$

$$
S_{1}<S_{2} .
$$

c) $a=10 \text{~cm}, b=2 \text{~cm}$;

$$
\begin{aligned}
& a=5 \text{~cm}, b=4 \text{~cm} \\
& p_{1}=18 \text{~cm} \\
& S_{1}=20 \text{~cm}^{2}
\end{aligned}
$$

$$
\begin{aligned}
& p_{1}=24 \text{~cm} ; \\
& S_{1}=20 \text{~cm}^{2}
\end{aligned}
$$

$$
S_{1}=S_{2} .
$$

Thus, we have three variants of the relationships between the areas of the rectangles.

Answer: the area of one rectangle can be greater than, less than, or equal to the area of another rectangle.