Question: I'm having some trouble making sense of this.

$\sqrt{\dfrac 12 \operatorname{in.}^2} =
\dfrac{1}{\sqrt 2} \operatorname{in.} =
\dfrac{1}{\sqrt 2} \operatorname{in.} \times \dfrac{\sqrt 2}{\sqrt 2} =
\dfrac{\sqrt 2}{2} \operatorname{in.}$

My question is in regards to the first two steps of the equation. How does one remember that taking the square root of a multi term requires you to distribute the square root to both terms? Is there a more intuitive explanation to this?

Like I know that:
$\sqrt{4\*4}$ is 2\*2 but is there a more intuitive way to remember this?

Answer: 
The way I think about it is that square root is no different to the power of one half and powers are multiplication in which the order doesn't matter. E.g.

$$\sqrt{4\times4}=\left(4\times4\right)^{\frac{1}{2}}=4^\frac{1}{2}\times4^\frac{1}{2}=\sqrt{4}\times\sqrt{4}=2\times2$$

Although all that happens in head automatically.
Completely formally, if $x$ and $y$ are nonnegative real numbers and $\alpha$ is a real number, then

$$(xy)^\alpha = x^\alpha \; y^\alpha$$

So
\begin{align}
 \sqrt{\dfrac 12 \operatorname{in.}^2}
 &= \left( \dfrac 12 \operatorname{in.}^2 \right)^{1/2} \\
 &= \left( \dfrac 12 \right)^{1/2} (\operatorname{in.}^2)^{1/2} \\
 &= \sqrt{ \dfrac 12} \; \sqrt{\operatorname{in.}^2} \\
 &= \dfrac{1}{\sqrt 2} \; \operatorname{in.}
\end{align}