Wednesday, March 2, 2016

Geometry Problem 1194: Equilateral Triangle, Perpendicular, Distance, Side, Metric Relations

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1194.

Math: Geometry Problem 1194: Equilateral Triangle, Perpendicular, Distance, Side, Metric Relations

5 comments:

  1. Draw MN middle line of ADEC (= 12)Apply theorem of pythagore MNB, ADB, BEC
    we get x=14

    ReplyDelete
  2. Draw CF ⊥ DA ..( F on DA)
    Let z= x^2
    Applying Pythagoras’s theorem in triangles ABD, BCE and ACF we get
    BE^2=x^2-121=z-121
    BD^2=z-169
    AC^2=CF^2+AF^2 => (BE+BD)^2+4=z
    Or 2.BE.BD=z-BE^2-BD^2-4= 286-z
    4.BE^2.BD^2=4(z-121)(z-169)=(286-z)^2
    Simplify this equation we get 3.z^2-588.z= 0
    We get 2 solutions z= 0 and z= 196
    Only z=196 is acceptable so x= 14

    ReplyDelete
    Replies
    1. See below for geometry solution of problem 1194

      http://s30.postimg.org/urgsbgadt/pro_1194.png
      Draw circles diameters BA and BC
      Draw altitudes BG of tri. ABC and GH of triangle DGE
      Observe that H is the midpoint of DE => DGE is isosceles tri.
      ∡ (DGE)= ∡ (DGB)+ ∡ (BGE)= ∡ (DAB)+ ∡ (BCE)= 180-120= 60 degrees
      So DGE is an equilateral tri.
      HG= ½( DA+CE)= 12= DE. √ (3)/2
      So DE= 8. √3=CF
      In right triangle ACF we have x^2= AF^2+CF^2= 4+(8√3)^2=196 => x= 14

      Delete
  3. DB + BE = √(x²-(13-11)²)
    √(x²-13²) + √(x²-11²) = √(x²-(13-11)²)
    x = 14

    ReplyDelete
  4. Sumith Peiris has left a new comment on your post "Geometry Problem 1194:
    Equilateral Triangle, Perpe...":

    Applying Pythagoras 3 times we see

    DE = sqrt(x^2-13^2) + sqrt(x^2-11^2)=sqrt(x^2-2^2)

    One can solve this equation by algebra but the equation becomes somewhat
    unwieldy

    By trial and error x = 14 (check: sqrt27 + sqrt75 = sqrt(27 + 75 +
    2Xsqrt(27X75)= sqrt(102 + 2 X 45)=sqrt192

    ReplyDelete