Irrational number approximations. Game on.

$6^{2} = 36$ and $7^{2} = 49\textsf{.}$ Since $36 < 40 < 49\textsf{,}$ $\sqrt{40}$ is between $6$ and $7\textsf{.}$

$5^2 = 25 < 27 < 36 = 6^2$, so $\sqrt{27}$ is between $5$ and $6$; the larger is $6$.

$9^2 = 81$ is very close to $83$ and $\sqrt{83} \approx 9.1$, so the closest integer is $9$.

$\sqrt{10} \approx 3.16$, so $\sqrt{10} + 2 \approx 5.16$; the integer part is $5$.

$\pi \approx 3.14159$.

$\sqrt{2} \approx 1.414$.

$\sqrt{3} \approx 1.732$.

$\sqrt{5} \approx 2.236$.