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Question Number 12148 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17

OA=OC=OD=OB=R,OE=((√2)/2)  BE⊥ED,FG⊥CD,FH⊥AB,∡OEB=∡OED.  find:  HE  and EG in term of : R

OA=OC=OD=OB=R,OE=22BEED,FGCD,FHAB,OEB=OED.find:HEandEGintermof:R

Commented by mrW1 last updated on 14/Apr/17

no unique solution

nouniquesolution

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 15/Apr/17

please help me by sending at last one  solution.thank you so much.

pleasehelpmebysendingatlastonesolution.thankyousomuch.

Commented by mrW1 last updated on 15/Apr/17

BE=ED=(((√(4R^2 −1))−1)/2)  (see Q12127)  but H can be any position on BE, and  G can be any position on ED.  I think in your question a condition  is missing.

BE=ED=4R2112(seeQ12127)butHcanbeanypositiononBE,andGcanbeanypositiononED.Ithinkinyourquestionaconditionismissing.

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 15/Apr/17

hello mrW1. HE and EG are wanted.  position of H and G are such that   alwyes OE=((√2)/2),∠OED=∠OEB.

hellomrW1.HEandEGarewanted.positionofHandGaresuchthatalwyesOE=22,OED=OEB.

Commented by mrW1 last updated on 15/Apr/17

you can only determine EB and ED,  but not EH and EG. to determine   EH and EG you need an additional  condition, e.g. F, E and O are collinear.

youcanonlydetermineEBandED,butnotEHandEG.todetermineEHandEGyouneedanadditionalcondition,e.g.F,EandOarecollinear.

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 15/Apr/17

F alweys is on circle O.O(−0.5,−0.5).  equation of circle can help us.  if F,E,O be collinear,we have special  case that gives us:EB=ED.

FalweysisoncircleO.O(0.5,0.5).equationofcirclecanhelpus.ifF,E,Obecollinear,wehavespecialcasethatgivesus:EB=ED.

Commented by mrW1 last updated on 15/Apr/17

please check the question once again.  the points H and G are free on line  EB and ED.

pleasecheckthequestiononceagain.thepointsHandGarefreeonlineEBandED.

Commented by mrW1 last updated on 15/Apr/17

Commented by mrW1 last updated on 15/Apr/17

(EH+(1/2))^2 +(EG+(1/2))^2 =R^2   there are infinite solutions for EH and EG.

(EH+12)2+(EG+12)2=R2thereareinfinitesolutionsforEHandEG.

Commented by chux last updated on 15/Apr/17

please which app do you use in   drawing this? n how is it used?.  thanks

pleasewhichappdoyouuseindrawingthis?nhowisitused?.thanks

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 15/Apr/17

hello. Geogebra app.

hello.Geogebraapp.

Commented by chux last updated on 15/Apr/17

thanks boss

thanksboss

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