computerages Posted January 18, 2007 Share 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? Link to comment Share on other sites More sharing options...
NeonBlack Posted January 18, 2007 Share 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. Link to comment Share on other sites More sharing options...
Bignose Posted January 18, 2007 Share 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. Link to comment Share on other sites More sharing options...
CPL.Luke Posted January 18, 2007 Share Posted January 18, 2007 yeah that simpifies down to (root 3)/4 s Link to comment Share on other sites More sharing options...
John Cuthber Posted January 27, 2007 Share 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. Link to comment Share on other sites More sharing options...
hotcommodity Posted January 28, 2007 Share Posted January 28, 2007 Should be [sqrt(3)/4]*s^2. http://www.mathwords.com/a/area_equilateral_triangle.htm Link to comment Share on other sites More sharing options...
D H Posted January 28, 2007 Share 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] Link to comment Share on other sites More sharing options...
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