See complete Problem 212 at:
gogeometry.com/problem/p212_triangle_120_equilateral_area.htm
120 degrees triangle, equilateral triangle, areas. Level: High School, SAT Prep, College geometry
Wednesday, December 24, 2008
Elearn Geometry Problem 212: 120 Degree Triangle, Areas
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Sa= (root3)*a^2/4
ReplyDeleteSb= (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
s=(root3)/4*ac
= (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.
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.
ReplyDeleteThis yields S=sqrt(Sa.Sc) and equations 1 and 2. QED