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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