How can area be measured in Radians? In this question: Any smart ideas on finding the area of this shaded region?, the area turns out to be $90−18.75 \pi - 25 \cdot \arctan(\frac{1}{2})$, which is $\approx 19.5$ Radians, according to wolfram alpha.
Amazingly enough (because I'm awful at math), I actually came to the same number using a very convoluted integration process (finding the area of the smaller circle segment by: making the diameter coincide with the $x$ axis, then finding the distance between the center of the circle and the line segment so that I could lower $y = \sqrt{-x(x-10)}$ by $\sqrt{5}$ to make the line segment coincide with the $x$ axis, solving for $0$ to find the $y$ intercepts (integration bounds), etc...(after that it's easy).
However, my assumption when I did this integration process was that my units were squared units of whatever measurement units I would choose to use (or be given) in this situation, for example squared cm, squared inches...squared anything, as I thought it had to be for area.
So my question is: How can the area be in radians, which are non-squared units? How does that even make sense? This almost sounds as if radians are a measure of area, essentially defined as how much area is in a sector of $x$ radians, this is very confusing.

The area isn't in radians. Wolfram Alpha just automatically assumes you want the answer in radians because there's an arctan in there.
It is true that the result of the arctan has to be "interpreted in radians", in the sense that we're taking $\arctan\frac 1 2$ to be the radian measure of the unique angle between $-\pi$ and $\pi$ such that the corresponding line has a slope of $\frac 1 2$. But, whatever. The point is that the result of that arctan is a number, and the value of that entire expression is a number, and that number is equal to the area of the figure in squared units of whatever is the unit of measurement was implicitly used in the problem.
From a dimensional analysis perspective, radians are "dimensionless", and you'll probably find if you look at the calculation that that $25$ has units of length squared.
