Tom Clarke and James Frederick Well from Queen Mary's Grammar School, Walsall and Shu Cao from Oxford High School gave this solution for a convex quadrilateral. Can you see how to adapt the solution for the case of the arrow shaped quadrilateral?

Suppose there is a convex quadrilateral $\mathrm{ABCD}$, the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ cross each other at $O$. The angle between $\mathrm{AO}$ and $\mathrm{BO}$ is $\theta$ degrees, the angle between $\mathrm{DO}$ and $\mathrm{CO}$ is the same. The angle between $\mathrm{AO}$ and $\mathrm{DO}$ is $180-\theta$ degrees, the angle between $\mathrm{BO}$ and $\mathrm{CO}$ is the same.

The area of triangle $\mathrm{AOB}$ is $\frac{1}{2}\mathrm{AO}\times \mathrm{BO}\sin \theta$.
The area of triangle $\mathrm{AOD}$ is $\frac{1}{2}\mathrm{AO}\times \mathrm{DO}\sin (180-\theta )=\frac{1}{2}\mathrm{AO}\times \mathrm{DO}\sin \theta$.
The area of triangle $\mathrm{DOC}$ is $\frac{1}{2}\mathrm{DO}\times \mathrm{CO}\sin \theta$.
The area of triangle $\mathrm{BOC}$ is $\frac{1}{2}\mathrm{BO}\times \mathrm{CO}\sin (180-\theta )=\frac{1}{2}\mathrm{BO}\times \mathrm{CO}\sin \theta$.
The area of the quadrilateral is the sum of these four triangles.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{\mathrm{Area}}& =\frac{1}{2}\mathrm{AO}\times \mathrm{BO}\sin \theta +\frac{1}{2}\mathrm{AO}\times \mathrm{DO}\sin \theta +\frac{1}{2}\mathrm{DO}\times \mathrm{CO}\sin \theta +\frac{1}{2}\mathrm{BO}\times \mathrm{CO}\sin \theta \\ \multicolumn{1}{c}{}& =\frac{1}{2}[\mathrm{AO}(\mathrm{BO}+\mathrm{DO})+\mathrm{CO}(\mathrm{DO}+\mathrm{BO})]\sin \theta \\ \multicolumn{1}{c}{}& =\frac{1}{2}(\mathrm{AO}\times \mathrm{BD}+\mathrm{CO}\times \mathrm{BD})\sin \theta \\ \multicolumn{1}{c}{}& =\frac{1}{2}\mathrm{AC}\times \mathrm{BD}\sin \theta \end{array}}$

So we have proved that for a convex quadrilateral the area of the quadrilateral is given by half the product of the lengths of the diagonals multiplied by the sine of the angle between the diagonals.