# How can the area of a square be less than its length?

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?

• You are trying to compare numerical values with different units – glowstonetrees Apr 14 '18 at 1:20
• i.e. there is nothing wrong with your reasoning, just that there is no meaning in comparing the values $0.001$ and $1/10^6$ since they have different units – glowstonetrees Apr 14 '18 at 1:23
• Related (maybe duplicate) math.stackexchange.com/questions/2635994/… – pjs36 Apr 14 '18 at 1:24
• Yes, the area of the whole stuck is not in fact $1/10^6 \text{m}$, but instead $1/10^6 \text{m}^2$, which explains a lot! – Robert Lewis Apr 14 '18 at 1:28
• The area is NOT $\frac 1{10^6} m$ but $\frac 1{10^6}m^2$ and as $1 mm = \frac 1{10^3}m$ then then unit of $1 mm^2 = \frac 1{10^6}m^2$ so that is no problem. – fleablood Apr 14 '18 at 2:49

## 2 Answers

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...

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.