A quadrilateral with no acute angels
2 answers:
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The artistic crop isn't helpful; it cuts off some vertex names.
The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn. We have right triangle ELH, right angle L, so
EH² = HL² + EL²
EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339
Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.
Answer: 4.25
