### Instruction: Evaluate $\lfloor{\sqrt{12}}\rfloor^2$.### Response: 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 $\sqrt{12}$.
I can estimate that $\sqrt{12}$ is between 3 and 4, since $3^2 = 9$ and $4^2 = 16$.
Therefore, $\lfloor{\sqrt{12}}\rfloor = 3$.
To find $\lfloor{\sqrt{12}}\rfloor^2$, I just need to square 3.
I know that $3^2 = 9$.