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
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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)
Ajit:ajitathle@gmail.com
The use of Al-Kashi's theorem or cosine rule and trigonometry can be avoided by sticking to strictly euclidean concepts.
ReplyDeleteAs <B=60°, if we let G be the foot of the altitude from A to BC, ABG is half an equilateral triangle and we have BG=AB/2=c/2.
Applying Euclid's propositions 13 of Book 2, in ABC we have b^2=a^2+c^2-2.BG.BC or b^2=a^2+c^2-a.c (1).
Any triangle with <B=60° will have an area of sqrt(3).a.c/4 (derived from Pythagore).
Using this formula for S, Sa, Sb and Sc and (1) yields the equations S=sqrt(Sa.Sc), Sb=Sa+Sc-S and Sb=Sa+Sc-sqrt(Sa.Sc) QED
With normal notation, AB=c, AC=b and BC=a
ReplyDeleteBy cosine law, b^2=a^2+c^2-2ac(cos60)
b^2=a^2+c^2-ac
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Sc=c^2(sin60)/2
Sa=a^2(sin60)/2
S=acsin(60)/2
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(1)
Sb=b^2(sin60)/2
=(a^2+c^2-ac)*sin60/2
=Sa+Sc-S
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(2)
It is easy to see S=sqrt(SaSc)
By replacing the S in the equation obtained in (1)
Sb=Sa+Sc-sqrt(SaSc)