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

## Wednesday, December 17, 2008

### Elearn Geometry Problem 211: 60 Degree Triangle, Areas

Labels:
60,
area,
congruence,
equilateral,
triangle

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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

ReplyDeleteNow 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)

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