Container Cuboid House

The painting area of this cuboid house transformed from a container is 70 square meters. Here is a detailed analysis:

I. Basic Dimensions of the House and Painting Scope The house has a cuboid structure, with a length of 6 meters, a width of 4 meters, and a height of 3 meters. The painting scope includes the four walls and the top, meaning the total area of five surfaces needs to be calculated. Among them, the top area is length × width = 6 × 4 = 24 square meters; the areas of the two front and back side walls are the same, each being length × height = 6 × 3 = 18 square meters, totaling 36 square meters; the areas of the two left and right side walls are the same, each being width × height = 4 × 3 = 12 square meters, totaling 24 square meters. Therefore, the total area of the five surfaces is 24 (top) + 36 (front and back side walls) + 24 (left and right side walls) = 84 square meters.

II. Deduction of Door and Window Areas It is clearly stated in the problem that the total area of the doors and windows of the house is 14 square meters. Since the door and window parts do not need to be painted, they should be deducted from the total area. Therefore, the actual area to be painted is 84 square meters (total area) - 14 square meters (door and window area) = 70 square meters.

III. Verification of Key Calculation Logic

1. Accuracy of area calculation: The surface area formula for a cuboid is 2 × (length × width + length × height + width × height), but in this problem, only the top and four walls need to be calculated, so the formula is simplified to length × width + 2 × (length × height + width × height). After substituting the data, the results are consistent.
2. Consistency of units: All dimensions are in meters, and the area is in square meters. There are no unit conversion errors in the calculation process.
3. Application of problem conditions: The deduction is strictly based on the door and window area given in the problem, without introducing additional assumptions or data.

IV. Precautions in Practical Applications If the house has other structures (such as eaves, decorative lines, etc.) or painting process requirements (such as two coats of painting, edge and corner treatment, etc.), the actual painting area may increase. However, this problem only calculates the theoretical value based on the given conditions.