Question

Triangle D E F is shown with point P at the center. Lines are drawn from each point of the triangle to point P. Line segments are drawn from point P to the sides of the triangle to form right angles and line segments P H, P J, and P G. The length of F J is 3 x minus 1, the length of J E is x + 3, the length of H E is 4 y minus 3, and the length of H D is 9.
Given that point P is equidistant from the vertices of ΔDEF, what is EF?



EF =

Answer 1

Because P is equidistant from the vertices, P is the circumcenter, so a perpendicular from P to a side bisects that side. Thus J is the midpoint of EF, so FJ = JE.

Set 3x − 1 = x + 3 → 2x = 4 → x = 2. Then FJ = 3(2) − 1 = 5, JE = 5, so EF = FJ + JE = 10.

EF = 10