Rooted via 10

Answer: $4\sqrt7$, $5\sqrt5$ and $6\sqrt3$ are greater than $10$

$\begin{align}\left(3\sqrt{11}\right)^2 &= 3\sqrt{11}\times3\sqrt{11}\\

&=3\times3\times\sqrt{11}\times\sqrt{11}\\

&=9\times11 = 99\lt100\end{align}$

So $3\sqrt{11}\lt10$

$\left(4\sqrt7\right)^2 = 16\times7 = 70 + 42\gt100$ so $4\sqrt7\gt10$

$\left(5\sqrt5\right)^2 = 25\times5 = 125 \gt100$ so $5\sqrt5\gt10$

$\left(6\sqrt3\right)^2 = 36\times3 = 108\gt100$ so $6\sqrt3\gt10$

$\left(7\sqrt2\right)^2 = 49\times2 = 98 \lt100$ so $7\sqrt2\lt10$