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.
Let $a, b \geq 0$. Each non-negative real number has a unique non-negative square root. Let $x = \sqrt{a}$; let $y = \sqrt{b}$. Observe that 

\begin{align\*}
ab & = x^2y^2\\
 & = x \cdot x \cdot y \cdot y\\
 & = x \cdot y \cdot x \cdot y\\
 & = xy \cdot xy\\ 
 & = (xy)^2\\
 & = (\sqrt{a}\sqrt{b})^2
\end{align\*}
Taking square roots yields $\sqrt{ab} = \sqrt{a}\sqrt{b}$.