Because
$$\sqrt{2}=\sqrt{\frac{8}{7}}\sqrt{\frac{7}{4}}\approx\frac{31}{29}\times\frac{41}{31}=\frac{41}{29}$$

We may group the three numbers into one single expression
$$4+6724\times6728=(6727-1)^2$$
that can be written as
$$\left(\frac{6727-1}{2}\right)^2-\left(\frac{6724}{2}\times\frac{6728}{2}\right)=1 $$
or
$$\left(\frac{7\times31^2-1}{2}\right)^2-2\left(2\times 29 \times 41\right)^2=1 $$

This is Pell's equation
$$X^2-dY^2=1$$

with $X=\frac{7\times31^2-1}{2}$, $Y={2\times29\times41}$ and $d=2$,

so the corresponding approximation to $\sqrt{2}$ is given by
$$\sqrt{2}\approx\frac{7\times31^2-1}{4\times29\times41}=\frac{(31\sqrt{7}+1)(31\sqrt{7}-1)}{4\times29\times41}=\frac{3363}{2378}$$
which is the tenth convergent of the continued fraction expansion.

Factoring the numerator shows that the square in 6727 is related to $\sqrt{7}$, as in the answer by wythagoras.

A simpler example is given by

$$\begin{align}98&=2\times7^2\\
99&=11\times3^2\\
100&=1\times10^2
\end{align}$$
with
$$99^2-98\times100=1$$

and $$\sqrt{2}\approx\frac{11\times3^2}{7\times10}=\frac{99}{70}$$

Approximating $\sqrt{2}$ with the sixth convergent explains the squares of $7$ and $10$, but we also need
$$\sqrt{\frac{11}{1}}\approx\frac{10}{3}$$
and / or
$$\sqrt{\frac{11}{2}}\approx\frac{7}{3}$$
to justify the square of $3$.

In this example, the approximation for $\sqrt{2}$ can be obtained by direct multiplication of the approximations for $\sqrt{\frac{2}{11}}$ and $\sqrt{\frac{1}{11}}$, but this is not the case in the example from the question.

$$\sqrt{2}=11\sqrt{\frac{2}{11}}\sqrt{\frac{1}{11}}\approx11\times\frac{3}{7}\times\frac{3}{10}=\frac{99}{70}$$

However, dividing the approximations implied by the equations involving number $7$, the convergent $\sqrt{2} \approx \frac{41}{29}$ is obtained.

$$\sqrt{2}=\sqrt{\frac{8}{7}}\sqrt{\frac{7}{4}}\approx\frac{31}{29}\times\frac{41}{31}=\frac{41}{29}$$