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In NJ Wildeberger's book (which I don't see much use in citing, since its an offline material, so instead I will provide a link to his key points further down) he argues that "Quadrance" can be just as useful as "Distance". He defines Quadrance as: "A squared plus B squared" & "(((C - A)^2)+((B - D)^2)) given lines 'AB' and 'CD'. More info here: " https://www.quora.com/Is-N-J-Wildberger-a-joke-or-a-genius-when-he-claims-that-mathematics-in-its-current-form-is-a-hoax " & here: " https://en.m.wikipedia.org/wiki/Rational_trigonometry " . Now knowing quadrance fills the shoes of distance; what fills the shoes of area? And how can it be converted back into its 'non-rational' equivalent?

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    $\begingroup$ You need to give more context. According to wikipedia, this is a relatively obscure topic in mathematics and I think that "quadrea" is not even part of the standard theory of rational trigonometry. $\endgroup$ Dec 22, 2017 at 10:31
  • $\begingroup$ @Raskolnikov Okay, is it alright if I spend a couple hours thinking about how to reframe it in the clearest possible context? Or should I just delete my question and try all over again? (I'm new) $\endgroup$
    – user179283
    Dec 22, 2017 at 10:45
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    $\begingroup$ I think you can rework the question. That will bump it. $\endgroup$ Dec 22, 2017 at 10:47
  • $\begingroup$ I have source material I could cite, but I made the mistake of assuming that it was common knowledge $\endgroup$
    – user179283
    Dec 22, 2017 at 10:47
  • $\begingroup$ Alright; I will re-work it, but due to limited char space I'm gonna need to put strategy into it $\endgroup$
    – user179283
    Dec 22, 2017 at 10:48

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The definition of quadrea is $ (Q_1+Q_2+Q_3)^2-2(Q_1^2+Q_2^2+Q_3^2).\;$ If $\;d_21^2=Q_1,\;$ $d_2^2=Q_2,$ $d_3^2=Q_3$, then it is $(d_1+d_2+d_3)(-d_1+d_2+d_3)(d_1-d_2+d_3)(d_1+d_2-d_3)$ and by Heron's formula this is $16$ times the square of the area of the triangle.

This can be checked by looking at an equilateral triangle side $d$ whose area squared is $\frac3{16}d^4.$

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  • $\begingroup$ Okay, to make sure my really base level stuff is still on point: to [conventionally] measure the area of a parrallelogram: you take all 4 points, measure their distance in an 'X'-shaped configuration (ie: measure its diagonals); then you divide the longest diagonal by 2, store that in variable "N"; and then multiply the shorter diagonal by variable "N". Correct? $\endgroup$
    – user179283
    Dec 22, 2017 at 20:41
  • $\begingroup$ If so, with quadrances what am I using/doing instead? Cos I only needed two distances... so, am I gonna just treat the quadrance based-solution like a "△" shaped point network, rather than the "V" shaped point network [that I used with distance on the diagonals to find the area in the previous comment]? I'm having trouble seeing which two points the third quadrance connects... ...I need help understanding this, so I can move past it $\endgroup$
    – user179283
    Dec 22, 2017 at 20:48

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