Why isn't the area of a square always greater than the length of one of its sides?

I think your intuition is failing you because you are trying to compare a 1-dimensional object (the length of a side) with a 2-dimensional object (the area of the interior of the square). You can fit loads of segments into a square of any size -- infinitely many, in fact! That comparison doesn't really mean anything.

On the other hand, here's a comparison that does make sense: Set a square of side length $s$ side-by-side with a rectangle whose sides are $s \times 1$. Now you are comparing area to area. The rectangle's area will fit inside the square if and only if $s>1$.


Because you're misunderstanding units. The first assumption you make is that a square with a side of $1$ has an area of $1$ - that assumption is incorrect.

A square with a side of $1000$ $m$ / $1$ $km$ / $0.001$ $Mm$ has an area of $1$ $km^2$, $1000000$ $m^2$, or $0.000001$ $Mm^2$ (square-Mega-meters), depending on how you chose to present it. It's all about presentation, not mathematical properties.

What you need to intuitively understand is that by doubling the length of the side of a square, you get 4 times the area. And by shrinking the side by half you shrink the area to a quarter, regardless of units.

Once you have that intuitive understanding, it will overrule your current understanding. Knowing that areas shrink "faster" than side lengths, it will be obvious that on a square with a side length of $1$ grok and an area of $1$ grikk, when you reduce the side length the area has to shrink faster than the side length - same is true for a square with a side length of $42$ gruk and an area of $42$ grakk: the area will shrink faster than the side length.


Take a square with dimensions $\cfrac 12 ft * \cfrac 12 ft $. The area would be $\cfrac 14$ $ft^2$. This makes perfect sense. I'll show you what I mean.

$\cfrac 12 ft* \cfrac 12 ft$ = $6 in*6 in$

The area is $36$ $in^2$ is equal to $1/4$ $ft^2$.

You can always take a square with the length of sides $x$ being $0<x<1$, but you can convert this $x$ to another unit $>0$. Therefore, the area would now make sense.

Tags:

Geometry

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