FERMAT 2010 Problem 19

Past Contests FERMAT FERMAT - 2010
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1

In the diagram, PQ=QR=RS=SP=SQ=6 and PT=RT=14. The length of ST is

Image

Answer Choices
A. 4\sqrt{10}-3
B. 11
C. 7\sqrt3-3
D. 10
E. \sqrt{232-84\sqrt3}