computerages Posted January 18, 2007 Posted January 18, 2007 hi, While solving a calculus problem today related to "cross sections" I derived a formula to calculate the area of an equilateral triangle. So I was just curious whether the formula I came up with is already in use or it's....? The formula: [math]A(s)=\frac{s}{2}\sqrt{s^2 - \frac{s^2}{4}}[/math] What do you guys think?
NeonBlack Posted January 18, 2007 Posted January 18, 2007 I think this is pretty impractical. It needs some serious simplification at least. Look, you have s^2 - 1/4 s^2. Why not just write 3/4 s^2? Then if you do that, you have the square root of a square.
Bignose Posted January 18, 2007 Posted January 18, 2007 And then, using the simplifications NeonBlack suggested, you get the result using trigonometry anyway... I was a little confused about your use of 's' as the variable at first, because I thought you were trying to replicate Heron's formula. Let s be the semiperimeter, s=0.5*(a+b+c) a,b,&c are the lengths of each of the three sides. Heron's forumla for the area of the triangle is then: Area = [ s*(s-a)*(s-b)*(s-c) ]^(1/2) I just saw that squareroot and the s and thought... now that isn't right, though if s is the length of the side of the equilateral triangle, then it is fine -- it does warrant simplification though.
John Cuthber Posted January 27, 2007 Posted January 27, 2007 Are you sure? It seems to me that an expression for an area should have a term which is second order in length. If I double the length of the side the area should quadruple rather than double.
hotcommodity Posted January 28, 2007 Posted January 28, 2007 Should be [sqrt(3)/4]*s^2. http://www.mathwords.com/a/area_equilateral_triangle.htm
D H Posted January 28, 2007 Posted January 28, 2007 Should be [sqrt(3)/4]*s^2. That is exactly the formula that computerages posted originally (just not reduced to simplest form): [math]A(s)=\frac{s}{2}\sqrt{s^2 - \frac{s^2}{4}} = \frac{s^2}{2}\sqrt{1 - \frac1 4} = \frac{s^2}{2}\sqrt{\frac3 4} = s^2\frac{\sqrt3}4[/math]
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