Wednesday, December 24, 2008

Elearn Geometry Problem 212: 120 Degree Triangle, Areas

Problem 211: Equilateral Triangle, Area

See complete Problem 212 at:

120 degrees triangle, equilateral triangle, areas. Level: High School, SAT Prep, College geometry


  1. Sa= (root3)*a^2/4
    Sb= (root3)*b^2/4
    Sc= (root3)*c^2/4
    S= 1/2*ac*Sin 120=(root3)*ac/4
    but by cosine rule
    b^2=a^2+c^2+ac so ac= b^2- a^2- c^2
    = (root3)/4[b^2-c^2-a^2]
    = Sb-Sc-Sa
    hence Sb=Sa+Sc+S -----(1)
    but S^2=Sa*Sc
    S= root of (Sa*Sc)
    substitute in equation 1
    we get Sb=Sa+Sc+ root of (Sc*Sa)
    hence solved.

  2. As in problem 211, the use of cosine rule and trigonometry can be avoided by using the foot G of the altitude from A to BC, noting that ABG is half an equilateral triangle, and applying both Pythagore to calculate the altitude AG, S, Sa, Sb and Sc,and applying proposition 13 of book II of Euclid's elements to triangle ABC.
    This yields S=sqrt(Sa.Sc) and equations 1 and 2. QED