In the figure below, C lies on the circumference of a circle with cent

Two ways.

First, the way I would do this problem if it were on an actual test. We can see that we will need to add two legs of 45-45-90 triangles and two hypotenuses (hyponetni?) of the same 45-45-90 triangles. Two of them are going to have values with no radical and two are going to have values containing \(\sqrt{2}\). Only one answer choice will fit that description.
Answer choice B.
That took maybe 15 seconds and there's essentially zero chance of making a silly mistake. No bonus points for doing the "real" math!!

Speaking of the "real" math, here's the second way.
The radius is 6, so CO is 6.
We are told that OA=OB and that BOA is 90 degrees. Since OA=OB, angle A = angle B. And since BOA is 90, A and B must both be 45.
If we connect CO, we create two 45-45-90 triangles.
CO=CB=6 and CO=CA=6.
That makes BO and AO both \(6\sqrt{2}\).
Adding AO+BO+BC+CA, we get \(6+6+6\sqrt{2}+6\sqrt{2}\)
\(12 + 12 \sqrt{2}\)
Answer choice B.