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amoodarya

  • one year ago

I know some formulas to find a triangle's area, like the ones below. 1. Is there any reference containing most triangle area formulas? 1. If you know more, please add them as an answer $$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$ Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix} a^2 & b^2 & c^2 \end{bmatrix}\begin{bmatrix} -1 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & -1 \end{bmatrix} \begin{bmatrix} a^2\\ b^2\\ c^2 \end{bmatrix}$$ [![triangle with three mutuaally tangent circles centr

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  1. amoodarya
    • one year ago
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    I know some formulas to find a triangle's area, like the ones below. 1. Is there any reference containing most triangle area formulas? 1. If you know more, please add them as an answer $$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$ Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix} a^2 & b^2 & c^2 \end{bmatrix}\begin{bmatrix} -1 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & -1 \end{bmatrix} \begin{bmatrix} a^2\\ b^2\\ c^2 \end{bmatrix}$$ [![triangle with three mutuaally tangent circles centred on the vertices][1]][1] Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles) $$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$ If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix} y_1 &z_1 &1 \\ y_2&z_2 &1 \\ y_3 &z_3 &1 \end{vmatrix}^2+\begin{vmatrix} z_1 &x_1 &1 \\ z_2&x_2 &1 \\ z_3 &x_3 &1 \end{vmatrix}^2+\begin{vmatrix} x_1 &y_1 &1 \\ x_2&y_2 &1 \\ x_3 &y_3 &1 \end{vmatrix}^2}$$ When we have 2-d coordinate $$ s=\frac{1}{2}\begin{vmatrix} x_a &y_a &1 \\ x_b &y_b &1 \\ x_c &y_c & 1 \end{vmatrix}$$ [![enter image description here][2]][3] In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by: $$ s=2R^2|sin(\frac{p}{2})sin(\frac{q}{2})sin(\frac{p-q}{2})|$$ \[s=\frac{abc}{4R}\\s=rp\]

  2. amoodarya
    • one year ago
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  3. amoodarya
    • one year ago
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  4. anonymous
    • one year ago
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    the 2nd and 3rd are identical, one just computes the altitude \(h_a\) using trig

  5. anonymous
    • one year ago
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    also why would you bother typing out all of this if you're just copying things from a mathworld article? http://mathworld.wolfram.com/TriangleArea.html

  6. anonymous
    • one year ago
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    the symmetrical form with a matrix is just because the expression under the radical is a quadratic form in \(a^2,b^2,c^2\), and in general a quadratic form \(Q\) is expressible as a matrix product \(x^T A x\) where \(A\) is a symmetric matrix

  7. amoodarya
    • one year ago
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    I see this page , you give I see it for the first time ! I collect them from my note (s) in many years

  8. anonymous
    • one year ago
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    hmm, well whatever source you collected these notes contains word-for-word copied sections? http://puu.sh/jalaS/288242df75.png http://puu.sh/jale2/7622d692ec.png

  9. amoodarya
    • one year ago
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    wow ! most of them was in "math passages in english " when I was b.s student it was many years ago ...

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