Thursday, February 19, 2015

Geometry Problem 1086: Square, Circle, Arc, Center, Metric Relations

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1086: Square, Circle, Arc, Center, Metric Relations.

3 comments:

  1. Since ∠OED=∠OHA, so OEDH concyclic.
    Thus ∠EOH=90°.

    Also, OE=OH=5.
    Hence, EH=5√2.

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  2. Rotate figure by 90° counter-clockwise around O. A goes to D and H goes to E. Hence OH=OE, EOH=90°, EH=5√2

    ReplyDelete
  3. Tr.s DOE and AOH are congruent SAS and so OH = OE

    So < AOH = < DOE = 67.5 and since < AOD = 90, HOE = 90

    Hence Tr. HOE is right isoceles and x = 5sqrt2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete