Is this true: $\mathrm{Cov}(X,Y) = \sqrt{\mathrm{Var(X)}}\sqrt{\mathrm{Var}(Y)}$? I just want to confirm whether this is correct or not:

The covariance of $X$ and $Y$ is equal to the standard deviation of $X$ times the standard deviation of $Y$.

or, in mathematical notation,
$$
\mathrm{Cov}(X,Y) = \sqrt{\mathrm{Var(X)}}\sqrt{\mathrm{Var}(Y)}
$$
 A: While cardinal has answered the question, I wanted to add a little something extra.
The Cauchy-Schwarz inequality shows that $\left|\operatorname{Cov}(X,Y)\right|\leq \sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}$, with equality if and only if $X$ and $Y$ are linearly dependent.  In general, you can take the ratio $\rho=\frac{\operatorname{Cov}(X,Y)}{ \sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}}$ which will be between $-1$ and $1$.  This gives the correlation between $X$ and $Y$.  It is $1$ when the variables are perfectly correlated (one goes up exactly when the other does), $-1$ when they are perfectly anti-correlated (one goes up exactly when the other goes down), $0$ when they are independent (knowing that one goes up tells you nothing about the behavior of the other), and in general gives a measure of the behavior when things are somewhere in between.
One of the first places where this comes up is least squares approximation.  .  While you can always do a least squares approximation to get a line of best fit for your data points, the correlation coefficient tells you whether your line of best fit is actually a good fit.  It will be near $0$ for data that is essentially random, and small in magnitude for data which is non-linear.
A: This is false in general. In fact, it is true if and only if $X=aY+b$ (almost surely) for some fixed constants $a \geq 0$ and $b \in \mathbb{R}$. That is $X$ and $Y$ must be positively linearly related for this to hold. Wikipedia also has a decent page on this.
For a counterexample to your statement, consider any two independent random variables $X$ and $Y$ each with strictly positive variance. Then,
$
\mathrm{Cov}(X,Y) = 0 \>,
$
but,
$
\sqrt{\mathrm{Var}(X)} \sqrt{\mathrm{Var}(Y)} > 0 \>.
$

A quick proof (and a slick one, I think; it's not my own) of the if and only if assertion uses the Cauchy–Schwarz inequality and goes as follows. Let $U$ and $V$ be random variables such that $\newcommand{\e}{\mathbb{E}}\e U^2 < \infty$ and $\e V^2 < \infty$. Then, $|\e UV|^2 < \e U^2 \e V^2$ if and only if $\e (t U - V)^2 > 0$ for all $t$. But, if $\e(t_0 U - V)^2 = 0$ for some $t_0$, then $t_0 U = V$ almost surely. Now, set $U = X - \e X$ and $V = Y - \e Y$ to get the desired conclusion. 
