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figure

In the above convex non intersecting quadrilateral we take AB=a, BC=b, CD=c, AD=d and $\angle ABC$ is $\theta$.

$$ Area of ABCD \\ = Ar\triangle ABC+Ar\triangle ACD \\ =\frac{1}{2}ab\sin\theta+\frac{1}{2}cd\sin\angle D \\ $$ Now the problem is to find $\sin\angle D$, using law of cosines we get: $$ AC^2=a^2+b^2-2ab\cos\theta \\ and AC^2=c^2+d^2-2cd\cos\angle D $$ Subtracting both equations from each other we get: $$ AC^2-AC^2=(a^2+b^2-2ab\cos\theta)-(c^2+d^2-2cd\cos\angle D), \\ a^2+b^2-c^2-d^2-2ab\cos\theta+2cd\cos\angle D=0, \\ \cos\angle D= \frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd} $$ Now using the trigonometric identity $ \sin\angle D= \sqrt{1-\cos^2\angle D} $ we get: $$ \sin\angle D= \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2} $$ And putting this into $ \frac{1}{2}ab\sin\theta+\frac{1}{2}cd\sin\angle D $ we get: $$ \frac{1}{2}\left[ab \sin \theta + cd \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2} \right] $$ Which I believe to be the area of any non intersecting convex quadrilateral. Please correct me if I'm wrong at any step. Also, do note that the a and b are the sides which are adjacent to the given angle, meaning that the lines a and b are the ones which make the given angle

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  • $\begingroup$ As a first check, what does it give you if you render $a=b=c=d=1$ and $\theta=90°$? $\endgroup$ Commented Jan 9 at 12:00
  • $\begingroup$ I get the area as 1 $\endgroup$ Commented Jan 9 at 12:05
  • $\begingroup$ Is that what it should be? What is that quadrilateral geometrically? $\endgroup$ Commented Jan 9 at 12:10
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    $\begingroup$ Yes. What you described is a square with sides 1 unit each so it's area is $ 1*1=1 unit^{2} $ $\endgroup$ Commented Jan 9 at 12:12

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The formulation appears to be correct. You already checked it for the case of a square.

You may want to consider the case where B is a concave vertex and the angle there is taken to be the reflex interior angle (>180°). In this way, recognizing that the sine of a reflex angle is negative, you find that the formula can be extended to include concave quadrilaterals too.

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  • $\begingroup$ Sure, I'll try to extend this formula to include concave quadrilaterals as well $\endgroup$ Commented Jan 9 at 15:16
  • $\begingroup$ I have one last question, is this formula that I've found an already known result or is it a completely new formula $\endgroup$ Commented Jan 9 at 16:10
  • $\begingroup$ Likely already known. The trigonometric laws from which it is derived have been around for centuries. $\endgroup$ Commented Jan 9 at 16:11
  • $\begingroup$ Ok thanks for telling, I'll still keep this post open for answers if in case anyone manages to find where this formula has been already found $\endgroup$ Commented Jan 9 at 16:17
  • $\begingroup$ Also, it should be noted that the sides a and b are the ones which form our given angle $\endgroup$ Commented Jan 10 at 8:13

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