Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Thursday, February 19, 2015
Geometry Problem 1086: Square, Circle, Arc, Center, Metric Relations
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Since ∠OED=∠OHA, so OEDH concyclic.
ReplyDeleteThus ∠EOH=90°.
Also, OE=OH=5.
Hence, EH=5√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
ReplyDeleteTr.s DOE and AOH are congruent SAS and so OH = OE
ReplyDeleteSo < AOH = < DOE = 67.5 and since < AOD = 90, HOE = 90
Hence Tr. HOE is right isoceles and x = 5sqrt2
Sumith Peiris
Moratuwa
Sri Lanka
For precision's sake, should'nt you have stated that both radiuses are equal?
ReplyDeleteIf this comment is for my proof, then let me say that I avoid the obvious steps for economy of words
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