The area of a square is calculated by squaring its length. For example, a square with a length of $2m$ yields an area of $4m^2$. However, if a square has a side length of $1mm$, then its area is $1mm^2$. In other words, it has a length of $0.001m$ but an area of $1/10^6$m, which is less than the length of any individual side. What is wrong with my reasoning here?
2026-03-27 00:04:46.1774569886
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How can the area of a square be less than its length?
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Perhaps it will help to think about this geometrically. Imagine drawing a square that is $1 \text{m} \times 1 \text{m}$. Then, imagine drawing a bunch of other squares that are each $1 \text{mm} \times 1 \text{mm}$ and cutting them out.
You can line up $1000$ of the smaller squares along one edge of the larger square. That is, the side length of the smaller square is one thousandth the length of the larger square.
On the other hand, you can place $1$ million of the smaller squares within the boundary of the larger square. That is, the area of the smaller square is one millionth the area of the entire square.
Looking at it this way, I think it is not so surprising the result you are seeing.
The units for area are square units. Area is $2$-dimensional, whereas length is $1$-dimensional. In a sense they aren't comparable...
A unit of length would be the length of a segment. For area, We have unit squares...
Similarly, volume is measured with unit cubes...
Then there is higher dimensional volume...