In the adjoining figure, $$AB = BC = CD$$. Point $$P$$ is the midpoint of $$AQ$$. If $$CR = 4$$ and $$QC = 12$$, then $$PQ$$ is equal to

Intersecting chords at $$C$$ give $$QC\cdot CR = BC\cdot CD$$, so the common length satisfies $$x^2=12\cdot4=48$$. Since $$P$$ is the midpoint of $$AQ$$, write $$AP=PQ=y$$ and $$AQ=2y$$. Using the two secants from $$A$$ gives $$AP\cdot AQ=AB\cdot AD$$, so $$2y^2=3x^2=144$$ and $$PQ=y=6\sqrt{2}$$. This value is not among the printed options, so the source treats the problem as a bonus and the answer letter is only a required placeholder.