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.

5 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

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  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
  4. For precision's sake, should'nt you have stated that both radiuses are equal?

    ReplyDelete
    Replies
    1. If this comment is for my proof, then let me say that I avoid the obvious steps for economy of words

      Delete