Wednesday, December 17, 2008

Elearn Geometry Problem 211: 60 Degree Triangle, Areas

Problem 211: Equilateral Triangle, Area

See complete Problem 211 at:
gogeometry.com/problem/p211_triangle_60_equilateral_area.htm

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

1 comment:

  1. We've S(a)+S(c)-S(b)= V3a^2/4 + V3c^2/4 -V3b^2/4 where V=square root. Hence, LHS=(V3/4)(a^2+c^2-b^2)--(1). By cosine rule in Tr. ABC,b^2=a^2+c^2-2ac*cos(60) or b^2=a^2+c^2 -2ac*(1/2) or ac=a^2+c^2-b^2. Using (1), S(a)+S(c)-S(b)= V3ac/4
    Now S=asin(60)*c/2 = V3ac/4. Thus, S(a)+S(c)-S(b)= S or S(b)= S(a)+S(c)-S
    We've S(a)=V3a^2/4 and S(c)=V3c^2/4.
    Hence, S(a)*S(c)=3a^2*c^2/16 or (S(a)*S(c))^(1/2)=V3ac/4 which is = S as shown earlier. Hence,
    S(b)= S(a) + S(c)-(S(a)*S(c))^(1/2)
    Ajit:ajitathle@gmail.com

    ReplyDelete