MHB NAP Yr 9 Example Test: Numeracy Q28 - 47.5 cm?

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Question 28 of the NAP Year 9 Numeracy test presents a discrepancy where one participant calculated an area of 47.5 cm², while others confirmed it as 46 cm². The correct approach involves calculating the areas of all rectangular faces and adding them to the given areas of triangular faces. A detailed breakdown of the calculations shows that the total area sums to 46 m², emphasizing the importance of unit conversion. Despite attempts to identify scenarios leading to the 47.5 cm² answer, no valid reasoning was found. Accurate calculations and unit awareness are crucial for solving such problems correctly.
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I get 46 cm². Can you post your work so we can see what you did wrong?
 

I get 46 too, but in meters and not in centimeters! (which makes a huge difference) (Sweating)

Areas of triangular faces are already given. We just need to find separate areas of all rectangular faces and add all triangular-face-areas to rectangular-face-areas in order to get the total area.

$$3+3+(5 \times 2.5)+(5 \times 2.5)+(5 \times 3) = 46 \text{m}^{2}$$

Also, I tried many experiments to find a situation where someone gets confused and gets the answer as $$47.5$$ by mistake, but found no such situation!
 

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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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