Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one corner, the second carpet $6 \times 6$ - in the opposite corner, and the third carpet $5 \times 7$ - in one of the remaining corners (all dimensions are in meters).

Find the area of the part of the hall covered by carpets in three layers (give the answer in square meters).

![](https://cdn.mathpix.com/cropped/2024_05_06_df4c0436ab68970fab15g-10.jpg?height=337&width=1003&top_left_y=304&top_left_x=225)
Answer: 6.

Solution. We will measure all dimensions in meters and the area in square meters.

Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 units from the bottom side.

![](https://cdn.mathpix.com/cropped/2024_05_06_df4c0436ab68970fab15g-10.jpg?height=315&width=321&top_left_y=922&top_left_x=566)

The first carpet intersects this rectangle horizontally between the 5th and 8th meters from the left side of the square room, and vertically between the 4th and 6th meters from the top side. In the end, this results in a rectangle $2 \times 3$, the area of which is 6.