Use the diagram below for questions 12-14, in which H is the circumcenter of ( riangle BCD), (BC = 18) and (HD = 14).
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Since H is the circumcenter, it is equidistant from all the vertices of the square. Let R be the circumradius, i.e., the distance from H to any vertex of the square. We can use the Pythagorean theorem to find R. Let's consider the right triangle HBD, where HD = Show more…
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