Geometry Problem.
Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 1000.
Friday, April 4, 2014
Geometry Problem 1000. Scalene Triangle, Isosceles, Angle, 120 Degree, Midpoint, Distance, Equilateral, Congruence, Perpendicular, Metric Relations
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CD=AE=2GH=8, GH=4, JH=2.
ReplyDeleteBy problem 998, FH=4.
Hence, x=√12=2√3.
http://s22.postimg.org/iff0mclox/problem_1000.png
ReplyDeleteLet P is the midpoint of DE
As the result of problem 998 we have
Triangle DBC congruent to ABE=> AF=DC=8
And triangle PFH and FHG are equilateral
Since G and H are midpoints of AC and EC => GH= 1/2AE=4
In equilateral triangle FGH altitude FJ= GH.sqrt(3)/2= 2.sqrt(3)
Following on from my proof of Problem 998, FGH is equilateral too and is of side 4 so
ReplyDeletex = sqrt(16 - 4) = 2sqrt3
Sumith Peiris
Moratuwa
Sri Lanka
Tr. ABE and CBD are congruent (SAS) => AE=8
ReplyDeleteSince G and H are mid-points of AC and EC => GH//AE and GH=4 -----(1)
Similarly GF//CD and GF=4 -----------------(2)
As ABD is isosceles=> m(BDF)=30 and Tr.DBF is 30-60-90 and BF=DB/2.
Similarly BH=EB/2
Tr. BFH similar to ABE with each side of BFH half of corresponding sides of ABE
Hence HF=AE/2=4 ---------(3)
From (1) ,(2) & (3) GFH is an equilateral triangle with each side = 4 units
=> FH=2Sqrt(3)