Friday, April 4, 2014

Geometry Problem 1000. Scalene Triangle, Isosceles, Angle, 120 Degree, Midpoint, Distance, Equilateral, Congruence, Perpendicular, Metric Relations

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.

Online Geometry Problem 1000. Scalene Triangle, Isosceles, Angle, 120 Degree, Midpoint, Distance, Equilateral, Congruence, Perpendicular, Metric Relations

4 comments:

  1. CD=AE=2GH=8, GH=4, JH=2.
    By problem 998, FH=4.
    Hence, x=√12=2√3.

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  2. http://s22.postimg.org/iff0mclox/problem_1000.png

    Let 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)

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  3. Following on from my proof of Problem 998, FGH is equilateral too and is of side 4 so
    x = sqrt(16 - 4) = 2sqrt3

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. Tr. ABE and CBD are congruent (SAS) => AE=8
    Since 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)

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