DimaMazin Posted November 8, 2015 Posted November 8, 2015 Fatness of triangle=A/P2 A/P2=1/(a+b+c) We know that: When angles of triangle are 60,60.60 degrees then a=b=c=12/(3)1/2 When triangle angles are 45,45,90 then a=2 b=2 c=2*21/2 When angles are 30,30,120 then a=14 b=14 c=16*31/2 Can we make any actions(the actions should be the same for any triangle) with triangle angles to define a,b,c ?
DimaMazin Posted November 8, 2015 Author Posted November 8, 2015 What are A and P? Area and perimeter.
michel123456 Posted November 8, 2015 Posted November 8, 2015 We know that: When angles of triangle are 60,60.60 degrees then a=b=c So far, so good, but you cannot put an absolute value for a b c .
Strange Posted November 8, 2015 Posted November 8, 2015 Fatness of triangle=A/P2 A/P2=1/(a+b+c) That doesn't see right. That would mean A = P2 / (a+b+c) Which, if P is the perimeter means: A = (a+b+c)2 / (a+b+c) A = (a+b+c) Which is not right. 1
John Cuthber Posted November 8, 2015 Posted November 8, 2015 (edited) The assertion that A/P2=1/(a+b+c) can't be right by dimensional analysis. Area has units of length squared, perimeter has units of length So the LHS of the equation has no units (they cancel) but the RHS has units of reciprocal length. If you want to call that unitless parameter "fatness" that's not a problem, but the rest of the post fails. Defining the perimeter of a triangle doesn't tell you what the area is http://www.mathopenref.com/triangleareaperim.html Even if you fix one of the sides there are still generally infinitely many possible triangles with the same perimeter. (the exception is that if you fix one side to be half the perimeter, then the area is zero, but that's hardly helpful) Edited November 8, 2015 by John Cuthber
DimaMazin Posted November 8, 2015 Author Posted November 8, 2015 That doesn't see right. That would mean A = P2 / (a+b+c) Which, if P is the perimeter means: A = (a+b+c)2 / (a+b+c) A = (a+b+c) Which is not right. Excuse me Strange. But a,b,c aren't sides of triangle. They should be just derivatives from angles like factorial or something else. The assertion that A/P2=1/(a+b+c) can't be right by dimensional analysis. Area has units of length squared, perimeter has units of length So the LHS of the equation has no units (they cancel) but the RHS has units of reciprocal length. If you want to call that unitless parameter "fatness" that's not a problem, but the rest of the post fails. Defining the perimeter of a triangle doesn't tell you what the area is http://www.mathopenref.com/triangleareaperim.html Even if you fix one of the sides there are still generally infinitely many possible triangles with the same perimeter. (the exception is that if you fix one side to be half the perimeter, then the area is zero, but that's hardly helpful) m2=m2 All is correct. So far, so good, but you cannot put an absolute value for a b c . I do make for equilateral triangle.
Strange Posted November 8, 2015 Posted November 8, 2015 Excuse me Strange. But a,b,c aren't sides of triangle. They should be just derivatives from angles like factorial or something else. So what are they? What is a "derivatives from angle"? How does that relate to factorials (n * n-1 * n-2 *n-3 ...) or "something else" (whatever that is)?
swansont Posted November 8, 2015 Posted November 8, 2015 Excuse me Strange. But a,b,c aren't sides of triangle. They should be just derivatives from angles like factorial or something else. IOW, you defined NONE of the terms in your OP. And you haven't even defined them here. That's enough nonsense.
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