Evaluate $\lfloor{\sqrt{12}}\rfloor^2$.
I recognize that $\lfloor x \rfloor$ is the floor function, which rounds down any real number to the nearest integer. So, $\lfloor{\sqrt{12}}\rfloor$ is the largest integer that is less than or equal to $\sqrt{12}$. Therefore, $\lfloor{\sqrt{12}}\rfloor = 3$. To find $\lfloor{\sqrt{12}}\rfloor^2$, I just need to square 3. I know that $3^2 = 9$. 