The units of the variance $S^2 = \frac{1}{n-1}\sum (X_i = \bar X)^2$ are squared. (If the $X_i$ are in $cm$ then $S^2$ has units $cm^2$.)
Using the sample standard deviation gets back to the original units.
Thus, if $X_i$ are a random sample from a normal distribution, one can write a 95% confidence interval as $\bar X \pm t^* S/\sqrt{n},$
where $t^*$ cuts probability 2.5% from the upper tail of Student's t distribution with $n-1$ degrees of freedom.
Division by $n-1$ gives $E(S^2) = \sigma^2,$ where $\sigma^2$ is the
population variance. This means that $S^2$ is an unbassed estimator
of $\sigma^2.$ However, note that $E(S) \ne \sigma$; the bias
is negligible for large $n$.
Also, some kinds of inference about $\sigma^2$
(for example, a confidence interval for $\sigma^2$) use the
fact that $(n-1)S^2/\sigma^2 \sim Chisq(df=n-1),$ the chi-squared
distribution with $n - 1$ degrees of freedom.
The formula for the population variance is usually written
with a capital $N$, denoting the population size: $\sigma^2 = \frac{1}{N}\sum (X_i - \mu)^2,$ where $X_i$ are the population
elements. (There is no discussion of using $N - 1$ here because
there is typically no need to estimate $\sigma^2$.)
Note: There would have been nothing "wrong" with defining $S^2$ using $n$ in the
denominator, and some statisticians have (belatedly) recommended that. But
using $n-1$ is pervasive and changing the definition of $S^2$ now would
turn out to require many adjustments in various formulas and tables used in inference.
