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.
You simply need to know that the square root function distributes over multiplication, so $$\sqrt{abc\cdots} = \sqrt{a}\sqrt{b}\sqrt{c} \cdots$$