If a square and triangle with sides measuring at 1 unit, have circles

If you have NOOOOO idea how to do geometry, you still have at least a 50/50 and can probably get to something like 80/20.

Look at the thing. The horizontal distance between A and B is 1. The vertical distance is really small. We want an answer choice that's just barely more than 1. A, B, and D aren't close. We are down to two. If you don't have a clue what to do next, congrats on having a 50/50, since that's way better than a blind guess!!

If you use a little logic, you can see that we're going to need to apply the distance formula (or Pythagoras...same thing), and that's going to mean a square root. Depending on what question number you're on, how much time you have left, your overall pacing strategy, and how quickly you think you can work through the geometry (without making a careless mistake!), you could very well feel comfortable with C. You've done no math and have something like an 80/20!!

If you've got the geometry chops to actually work through the math, we can get that to 100/0.
Treat the bottom left of the diagram as the origin in a coordinate plane, so point B lies at (1.5, 0.5).
The triangle O-A-centerofbaseoftriangle is a 30-60-90. The longer leg is 1/2. So the shorter leg is (1/2)/\(\sqrt{3}\) = \(\sqrt{3}/6\)
Point A lies at (0.5, \(\frac{\sqrt{3}}{6}\)).
Making a right triangle between A and B, the horizontal leg is 1 and the vertical leg is \(\frac{1}{2}-\frac{\sqrt{3}}{6}=\frac{3-\sqrt{3}}{6}\).
\(d^2=1^2+(\frac{3-\sqrt{3}}{6})^2\)
\(d^2=1+(\frac{9-6\sqrt{3}+3}{36})\)
\(d^2=\frac{36}{36}+(\frac{12-6\sqrt{3}}{36})\)
\(d^2=\frac{48-6\sqrt{3}}{36}\)
\(d^2=\frac{8-\sqrt{3}}{6}\)
Take the square root of both sides.

Answer choice C.