1. The problem is to find the square root of 81. 2. The square root of a number $x$ is a value $y$ such that $y^2 = x$. 3. Here, we want to find $y$ such that $y^2 = 81$. 4. Since $9^2 = 81$, the square root of 81 is 9. 5. Note that both 9 and -9 squared give 81, but the principal (positive) square root is 9. Final answer: $\sqrt{81} = 9$