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
