7.2. A square with a side of $100 \mathrm{~cm}$ was drawn on the board. Alexei crossed it with two lines parallel to one pair of sides of the square. Then Danil crossed the square with two lines parallel to the other pair of sides of the square. As a result, the square was divided into 9 rectangles, and it turned out that the lengths of the sides of the central section were 40 cm and 60 cm. Find the sum of the areas of the corner rectangles.
Answer: 2400.

Solution 1: Without loss of generality, we will assume that the central rectangle has a width of 60 cm and a height of 40 cm. Let $x$ and $y$ be the width and height, respectively, of the lower left rectangle. Then the upper left rectangle has sides $x$ and $(60-y)$, the upper right rectangle has sides $(40-x)$ and $(60-y)$, and the lower right rectangle has sides $(40-x)$ and $y$. The total area of the corner rectangles is

$$
x y + 60 x - x y + 40 y - x y + 2400 - 60 x - 40 y + x y = 2400.
$$

Solution 2: Suppose Alexei not only drew the lines but also cut out the entire central part bounded by these lines, and then Daniil did the same. Then, from the original square, only a rectangle of 60 by 40 remains on one side. On the other hand, everything that remains is the rectangles in question. Therefore, the area of the corner rectangles is equal to the area of the remaining rectangle and is 2400.

Criteria: Only the answer - 0 points. Answer with verification - 1 point.

In the solution, it is explicitly assumed that the cuts were made along the grid lines: if the solution relies on this (for example, by enumeration) - no more than 3 points; if the solution does not use this - do not deduct points.