Post History
#5: Post edited
by
clemens
·
2026-05-20T07:46:54Z (4 months ago)
typofixes
Area of a (planar) quadrilateral using two opposite sides and four angles
- I guessed the following area formula for a (planar) quadrilateral using two opposite sides and four angles:(When $a=AB$ and $c=CD$,)
- $S = \frac{a^2}{2(\cot A + \cot B)} + \frac{c^2}{2(\cot C + \cot D)}$
- This formula can be applied to the quadrilateral satisfying $A+B \ne \pi$ and $C+D \ne \pi$, including convex, concave, and even self-intersecting cases.
I'd like to know the applicabile case , inapplicabile case and reasons for the validity of those cases. Are there any known papers or more elegant ways to derive this formula?- I guess there are many ways to explain those reasons. It greatly resembles $S = \frac{1}{2}(ab \sin B + cd \sin D)$. It seems that $b$ and $d$ can be determined using $a, c, A, B, C, D$.
- I guessed the following area formula for a (planar) quadrilateral using two opposite sides and four angles:(When $a=AB$ and $c=CD$,)
- $S = \frac{a^2}{2(\cot A + \cot B)} + \frac{c^2}{2(\cot C + \cot D)}$
- This formula can be applied to the quadrilateral satisfying $A+B \ne \pi$ and $C+D \ne \pi$, including convex, concave, and even self-intersecting cases.
- I'd like to know the applicable case, inapplicable case and reasons for the validity of those cases. Are there any known papers or more elegant ways to derive this formula?
- I guess there are many ways to explain those reasons. It greatly resembles $S = \frac{1}{2}(ab \sin B + cd \sin D)$. It seems that $b$ and $d$ can be determined using $a, c, A, B, C, D$.
#4: Post edited
by
Ryo TAKA
·
2026-05-19T23:28:39Z (4 months ago)
- I guessed the following area formula for a (planar) quadrilateral using two opposite sides and four angles:(When $a=AB$ and $c=CD$,)
- $S = \frac{a^2}{2(\cot A + \cot B)} + \frac{c^2}{2(\cot C + \cot D)}$
- This formula can be applied to the quadrilateral satisfying $A+B \ne \pi$ and $C+D \ne \pi$, including convex, concave, and even self-intersecting cases.
- I'd like to know the applicabile case , inapplicabile case and reasons for the validity of those cases. Are there any known papers or more elegant ways to derive this formula?
- I guess there are many ways to explain those reasons. It greatly resembles $S = \frac{1}{2}(ab \sin B + cd \sin D)$. It seems that $b$ and $d$ can be determined using $a, c, A, B, C, D$.
- I guessed the following area formula for a (planar) quadrilateral using two opposite sides and four angles:(When $a=AB$ and $c=CD$,)
- $S = \frac{a^2}{2(\cot A + \cot B)} + \frac{c^2}{2(\cot C + \cot D)}$
- This formula can be applied to the quadrilateral satisfying $A+B \ne \pi$ and $C+D \ne \pi$, including convex, concave, and even self-intersecting cases.
- I'd like to know the applicabile case , inapplicabile case and reasons for the validity of those cases. Are there any known papers or more elegant ways to derive this formula?
- I guess there are many ways to explain those reasons. It greatly resembles $S = \frac{1}{2}(ab \sin B + cd \sin D)$. It seems that $b$ and $d$ can be determined using $a, c, A, B, C, D$.
#3: Post edited
by
Ryo TAKA
·
2026-05-19T23:26:55Z (4 months ago)
#2: Post edited
by
Ryo TAKA
·
2026-05-19T23:26:27Z (4 months ago)
**Area of a (planar) quadrilateral using two opposite sides and four angles**- I guessed the following area formula for a (planar) quadrilateral using two opposite sides and four angles:(When $a=AB$ and $c=CD$,)
- $S = \frac{a^2}{2(\cot A + \cot B)} + \frac{c^2}{2(\cot C + \cot D)}$
- This formula can be applied to the quadrilateral satisfying $A+B \ne \pi$ and $C+D \ne \pi$, including convex, concave, and even self-intersecting cases.
- I'd like to know the applicabile case , inapplicabile case and reasons for the validity of those cases. Are there any known papers or more elegant ways to derive this formula?
- I guess there are many ways to explain those reasons. It greatly resembles $S = \frac{1}{2}(ab \sin B + cd \sin D)$. It seems that $b$ and $d$ can be determined using $a, c, A, B, C, D$.
- I guessed the following area formula for a (planar) quadrilateral using two opposite sides and four angles:(When $a=AB$ and $c=CD$,)
- $S = \frac{a^2}{2(\cot A + \cot B)} + \frac{c^2}{2(\cot C + \cot D)}$
- This formula can be applied to the quadrilateral satisfying $A+B \ne \pi$ and $C+D \ne \pi$, including convex, concave, and even self-intersecting cases.
- I'd like to know the applicabile case , inapplicabile case and reasons for the validity of those cases. Are there any known papers or more elegant ways to derive this formula?
- I guess there are many ways to explain those reasons. It greatly resembles $S = \frac{1}{2}(ab \sin B + cd \sin D)$. It seems that $b$ and $d$ can be determined using $a, c, A, B, C, D$.
#1: Initial revision
by
Ryo TAKA
·
2026-05-19T23:25:40Z (4 months ago)
Area of a (planar) quadrilateral using two opposite sides and four angles
**Area of a (planar) quadrilateral using two opposite sides and four angles**
I guessed the following area formula for a (planar) quadrilateral using two opposite sides and four angles:(When $a=AB$ and $c=CD$,)
$S = \frac{a^2}{2(\cot A + \cot B)} + \frac{c^2}{2(\cot C + \cot D)}$
This formula can be applied to the quadrilateral satisfying $A+B \ne \pi$ and $C+D \ne \pi$, including convex, concave, and even self-intersecting cases.
I'd like to know the applicabile case , inapplicabile case and reasons for the validity of those cases. Are there any known papers or more elegant ways to derive this formula?
I guess there are many ways to explain those reasons. It greatly resembles $S = \frac{1}{2}(ab \sin B + cd \sin D)$. It seems that $b$ and $d$ can be determined using $a, c, A, B, C, D$.