Problem:
A convex quadrilateral has sides $10$, $6$, $8$ and $2$ (in that order). If the angle between the diagonals is $π/4$, then find the area of the quadrilateral.
I tried applying the Law of Cosines for the angles in between the diagonals, but arrived at a weird relation.
Any hints?
PS: I don't know much Latex, so I had to use radians instead of degrees.
If distances from the intersection point of the diagonals to the vertices are $a, b, c, d$, here is the relations between them -
$a^2+b^2 - 2 ab \cos 135^0 = 64$
Please note the area of $\triangle AOB$ is $A_1 = \frac{1}{2} ab \sin 135^0 = - \frac{1}{2} ab \cos 135^0$
So, $a^2+b^2 + 4A_1 = 64$ ...(i)
Similarly, $c^2+d^2 + 4A_3 = 100$ ...(ii)
You also get,
$b^2+d^2 - 4A_2 = 36$ ...(iii)
$c^2+a^2 - 4A_4 = 4$ ...(iv)
Subtract (iii) from (i), (iv) from (ii) and then add. What do you get?