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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Details: Click on the figure below.
Problem 1333Let (ACE) is area triangle ACE then (ACE)=CE.AD/2=GH.AD/2=(ADH)=42.(KDH)=(KAH)-(ADH)=56-42=14. But EH//CG (AE perpendicular CG) so GH=CE.Is EDHG trapezium so x=(KEG)=(KDH)=14.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Name: AD=a, DG=b, S(AED)=m, S(EDK)=n(a²)/2=42+m, (b²)/2=x+n, (AE²)/2=56+m+n,Using AE²=a²+b² => x=14
Let AD = a and DG = bAEGH is concyclic and so Tr. AEH is right isosceles. So Tr.s AED and EFH are congruent and GH = a-b.S(AGH)-S(HEG) = 56-x = 1/2((a+b)(a-b) - b(a-b))So 56-x = 1/2a(a-b) = S(AEC) = 42Hence x = 56-42 = 14Sumith PeirisMoratuwaSri Lanka
Three triangle are similar.by pytagorian theoremx=56-42=14