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.
Well, we know that:
$$a=\sqrt a\sqrt a\tag1$$
almost by definition. Also:
$$a=\sqrt{aa}\tag2$$
for positive $a$. (From now on, I'll assume $a$ and $b$ are positive.)
So:
\begin{align}
\sqrt{ab}&=\sqrt{(\sqrt a\sqrt a)(\sqrt b\sqrt b)}\\
&=\sqrt{(\sqrt a)(\sqrt a)(\sqrt b)(\sqrt b)}\\
&=\sqrt{(\sqrt a)(\sqrt b)(\sqrt a)(\sqrt b)}\\
&=\sqrt{(\sqrt a\sqrt b)(\sqrt a\sqrt b)}\\
&=\sqrt a\sqrt b
\end{align}
I used equation $(1)$ for the first step, associativity and commutativity of multiplication for the next three steps, and then equation $(2)$ at the end.