Expanding on @gung's excellent answer:
In a simple linear regression the absolute value of Pearson's $r$ can be seen as the geometric mean of the two slopes we obtain if we regress $y$ on $x$ and $x$ on $y$, respectively:
$$\sqrt{\hat\beta_{y\,on\,x} \cdot \hat\beta_{x\,on\,y}} = \sqrt{\frac{\text{Cov}(x,y)}{\text{Var}(x)} \cdot \frac{\text{Cov}(y,x)}{\text{Var}(y)}} = \frac{|\text{Cov}(x,y)|}{\text{SD}(x) \cdot \text{SD}(y)} = |r|
$$$$\sqrt{{\hat{\beta}_1}_{y\,on\,x} \cdot {\hat{\beta}_1}_{x\,on\,y}} = \sqrt{\frac{\text{Cov}(x,y)}{\text{Var}(x)} \cdot \frac{\text{Cov}(y,x)}{\text{Var}(y)}} = \frac{|\text{Cov}(x,y)|}{\text{SD}(x) \cdot \text{SD}(y)} = |r|
$$
We can obtain $r$ directly using
$$r = sign(\hat\beta_{y\,on\,x}) \cdot \sqrt{\hat\beta_{y\,on\,x} \cdot \hat\beta_{x\,on\,y}}
$$$$r = sign({\hat{\beta}_1}_{y\,on\,x}) \cdot \sqrt{{\hat{\beta}_1}_{y\,on\,x} \cdot {\hat{\beta}_1}_{x\,on\,y}}
$$
or
$$r = sign(\hat\beta_{x\,on\,y}) \cdot \sqrt{\hat\beta_{y\,on\,x} \cdot \hat\beta_{x\,on\,y}}
$$$$r = sign({\hat{\beta}_1}_{x\,on\,y}) \cdot \sqrt{{\hat{\beta}_1}_{y\,on\,x} \cdot {\hat{\beta}_1}_{x\,on\,y}}
$$
Interestingly, by the AM–GM inequality, it follows that the absolute value of the arithmetic mean of the two slope coefficients is greater than (or equal to) the absolute value of Pearson's $r$: $$ |\frac{1}{2} \cdot (\hat\beta_{y\,on\,x} + \hat\beta_{x\,on\,y})| \geq \sqrt{\hat\beta_{y\,on\,x} \cdot \hat\beta_{x\,on\,y}} = |r| $$$$ |\frac{1}{2} \cdot ({\hat{\beta}_1}_{y\,on\,x} + {\hat{\beta}_1}_{x\,on\,y})| \geq \sqrt{{\hat{\beta}_1}_{y\,on\,x} \cdot {\hat{\beta}_1}_{x\,on\,y}} = |r| $$