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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.
Let y= FBWe have Triangle AOE similar to AFB => FA= 2.yWe have y^2+4y^2= 16 => y= 4/sqrt(5)Applying Ptolemy's theorem in quad. ADBD we haveAB.FD=DB.AF+FB.AD4.x= 2.sqrt(2).( 2y +y)So x= 3.y.sqrt(2)/2 Replace value of y in above we have x= 12.sqrt(10)/5
Thanks AntonioYes, The correct answer is 6.sqrt(10)/5.the error is in the last stepPeter
cos∠FOD = -sin∠FOB= -2 sin∠FAB cos∠FAB= -2×1/√5×2/√5= -4/5In ΔFOD, x² = 2²+2²-2×2×2cos∠FOD = 72/5x = 6√10 / 5
Observam ca AE=√5 (TP in trAOE) apoi CE.ED=FE.EA<=>1.3=√5.FE,teorema medianei in tr CFO ne da relatiaFE^2=[2(FO^2+FC^2)-CO^2]/4 si in final TP in tr CFD =>FO^2=CD^2-CF^2 =>x=4√21/5
Observam ca AE=√5 (TP in trAOE) apoi CE.ED=FE.EA<=>1.3=√5.FE,teorema medianei in tr CFO ne da relatiaFE^2=[2(FO^2+FC^2)-CO^2]/4 =>FE^2=8/5si in final TP in tr CFD =>FO^2=CD^2-CF^2 =>FO^2=16-8/5=>x=6√10 / 5
Thanks Ion, the second is the right answer.
Draw FG perp to CD.Triangles EFG, EAO are similar.So FG/AO = GE/EO =EF/AE.(i.e.) FG/2 = GE/1 = EF/√5.Note AE.EF = DE.EC = 3*1 = 3.(i.e.) √5*EF=3, EF = 3/√5.So FG=6/5, GE=3/5, DG = 3 + 3/5 =18/5So x^2 = DG^2 + FG^2 = 324/25 + 36/25 = 360/25.Hence x = 6√10 / 5
Choose O as origin, lines AOB, COD as coordinate axes.A=(-2,0),E=(0,1) imply equation of AE is x - 2y + 2=0.Eq of the circle is x^2 + y^2 = 4Solving these two equations, [x = 0, y = - 2] or [x = 6/5, y = 8/5]So F=(6/5 ,8/5). Already D = (0,-2)Hence DF^2 = 36/5 + 324/25 = 360/25, DF = 6√10 / 5
CF^2 = 16-x^2, AE • sqrt 5 and AD = sqrt 8Tr.s ECF and AED are similar so (16-x^2)/8 • 1/5 from which x • 6/5sqrt10Sumith PeirisMoratuwa Sri Lanka
Remark only: If G is the intersection of DF and AB, then S[DEG]=S[AEC].Happy New Year everybody!